Znzryb's Blog

Codeforces Round 1046 (Div. 2)——D. For the Champion

Posted on 2025-08-28 Edited on 2026-02-18

思路讲解

这种题目就是要多想，多写，想办法去除绝对值。

把式子写下来。

    ll dis=ask('U',mod);
    sort(all(A),[&](Point a,Point b){
        return 2*2*mod-a.x()-a.y() < 2*2*mod-b.x()-b.y();
    });
    Point a=A[0];
    // 知道了x+y是多少
    // mod*2+X+mod*2+Y-a.x()-a.()=dis
    ll XY=dis-2*2*mod+a.x()+a.y();
// #ifdef LOCAL
//     debug(XY);
// #endif
    ask('D',mod);
    ask('D', mod);
    ask('D',mod);
    ll dis2=ask('D', mod);
    sort(all(A),[&](Point a,Point b){
        return 2*2*mod-a.x()+a.y() < 2*2*mod-b.x()+b.y();
    });
    a=A[0];
    // X+2*mod-x+y-(Y-2*mod)=dis2
    // dis2+x-y-2*2*mod
    ll X_Y=dis2+a.x()-a.y()-2*2*mod;

AC代码

https://codeforces.com/contest/2136/submission/336079080

源代码

// by Gospel_rock
#include <algorithm>
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define PB push_back
#define EB emplace_back
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((long long) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define fsp(x) fixed<<setprecision(x)
#define minn(vec) *min_element(vec.begin(),vec.end())
#define maxx(vec) *max_element(vec.begin(),vec.end())
using namespace std;
#ifdef LOCAL
template<typename T> void _print(T x) { cerr << x; }
// 可以打印多层数组，但是格式不是很好
template<typename T> void _print(vector<T> v) { cerr << "[ "; for (T i : v) _print(i), cerr << " "; cerr << "]"; }
template<typename T> void _print(set<T> s) { cerr << "{ "; for (T i : s) _print(i), cerr << " "; cerr << "}"; }
template<typename T, typename U> void _print(pair<T, U> p) { cerr << "{"; _print(p.first); cerr << ", "; _print(p.second); cerr << "}"; }
template<typename T, typename U> void _print(map<T, U> m) { cerr << "{ "; for (auto &p : m) _print(p), cerr << " "; cerr << "}"; }
// 打印两层数组，格式好一些
template<typename T> void _print(vector<vector<T>> v){cerr<<"{\n";for (auto i : v) _print(i), cerr << "\n";cerr<<" }";}
#endif

using  ll = long long;using ull = unsigned long long;
using ld=double;using LD=long double;
using i128 = __int128;
using pdd = pair<ld,ld>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;using vb=vector<bool>;
using tll = tuple<ll,ll,ll>;using vii=vector<int>;
using vpll = vector<pll>;using vtll = vector<tll>;
using vdb = vector<ld>;using vc=vector<char>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
using CD=complex<double>;
static constexpr ll MAXN=(ll)1e6+10,INF=(ll)1e17+3; // 1e17+3是素数 1e18+9是素数
static constexpr ll mod=(ll)1e9; // 998244353;
static constexpr double eps=1e-8;const double pi=acos(-1.0);
ll N,M,Q,X,K,T,lT;

inline void __();
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    
    cin>>lT;
    while(lT--)
        __();
    return 0;
}
/*

*/
using Real = ll;
using Point = complex<Real>;
#define x() real()
#define y() imag()
Real cross(const Point &a, const Point &b) {
    return (conj(a) * b).imag();
} 
Real dot(const Point &a, const Point &b) {
    return (conj(a) * b).real();
}
Real mandis(const Point &a,const Point &b){
    return abs(a.x()-b.x())+abs(a.y()-b.y());
}
int sgn(Real x) {
    if (-eps < x && x < eps) return 0;
    return x < 0 ? -1 : 1;
}
namespace std
{
    bool operator <(const Point& a, const Point& b){
        return sgn(real(a) - real(b)) ? real(a) < real(b) : imag(a) <
               imag(b);
    }
    bool operator ==(const Point& a, const Point& b){
        if(!sgn(a.imag()-b.imag()) && !sgn(a.real()-b.real())){
            return true;
        }
        return false;
    }
    bool operator !=(const Point& a, const Point& b){
        if(!sgn(a.imag()-b.imag()) && !sgn(a.real()-b.real())){
            return false;
        }
        return true;
    }
    istream& operator>>(istream& in, Point& c) {
        Real r, i;
        in >> r >> i;
        c = Point(r, i);
        return in;
    }
}
// 采用两点式
struct Line{
    Point a,b;
    Line(){}
    Line(Point a,Point b):a(a),b(b) {}
    bool operator<(const Line&other) const{
        return a!=other.a?a<other.a:b<other.b;
    }
    bool operator==(const Line&other) const{
        return a==other.a && b==other.b;
    }
};
istream& operator>>(istream& in, Line& e) {
    Real a,b,c,d;
    in >> a >> b >> c >> d;
    e = Line(Point(a,b),Point(c,d));
    return in;
}
// p+tv=q+sw
// (p-q) x w+tv x w = 0
// t=-(p-q)x w/(vxw)
// 传入的是方向式  拿到的是两直线交点
Point getintersect(const Line &a,const Line &b){
    Point p=a.a, v=a.b-a.a,q=b.a,w=b.b-b.a;
    Point u=p-q;
    Real t=cross(w,u)/cross(v,w);
    return p+v*t;
}
bool isseginterline(Line seg,Line e){
    if(sgn(cross(e.a-seg.a, e.b-seg.a))*sgn(cross(e.a-seg.b, e.b-seg.b))>0){
        return false;
    }
    return true;
}
ll ask(char a,ll b){
    cout<<"? "<<a<<" "<<b<<endl;
    ll res;cin>>res;
    return res;
}
inline void __(){
    cin>>N;
    if(N==-1) exit(0);
    vector<Point> A;
    FOR(i,1,N){
        Point a;cin>>a;
        A.EB(a);
    }
    ask('R',mod);
    ask('R',mod);
    ask('U',mod);
    ll dis=ask('U',mod);
    sort(all(A),[&](Point a,Point b){
        return 2*2*mod-a.x()-a.y() < 2*2*mod-b.x()-b.y();
    });
    Point a=A[0];
    // 知道了x+y是多少
    // mod*2+X+mod*2+Y-a.x()-a.()=dis
    ll XY=dis-2*2*mod+a.x()+a.y();
// #ifdef LOCAL
//     debug(XY);
// #endif
    ask('D',mod);
    ask('D', mod);
    ask('D',mod);
    ll dis2=ask('D', mod);
    sort(all(A),[&](Point a,Point b){
        return 2*2*mod-a.x()+a.y() < 2*2*mod-b.x()+b.y();
    });
    a=A[0];
    // X+2*mod-x+y-(Y-2*mod)=dis2
    // dis2+x-y-2*2*mod
    ll X_Y=dis2+a.x()-a.y()-2*2*mod;
// #ifdef LOCAL
//     debug(X_Y);
// #endif
    ll X=(XY+X_Y)/2;
    ll Y=XY-X;
    cout<<"! "<<X<<" "<<Y<<endl;
}
/*
AC
https://codeforces.com/contest/2136/submission/336079080
1
1
0 0
? D 1000000000
1
? D 1000000000
1
? R 1000000000
1
? R 1000000000
4000000000
? U 1000000000
1
? U 1000000000
1
? U 1000000000
1
? U 1000000000
4000000000
! 0 0

*/

心路历程（WA，TLE，MLE……）

Treasure Hunt

Posted on 2025-08-28 Edited on 2026-02-18

思路讲解

The Doors 的想法，跑dij。

感觉可以套用这个做法，把距离都变成1，就是要开多少门。

AC代码

https://vjudge.net/solution/63304444

源代码

// by Gospel_rock
#include <algorithm>
#include <bits/stdc++.h>
#include <istream>
#include <ostream>
#include <queue>
#include <vector>
#define all(vec) vec.begin(),vec.end()
#define PB push_back
#define EB emplace_back
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((long long) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define fsp(x) fixed<<setprecision(x)
#define minn(vec) *min_element(vec.begin(),vec.end())
#define maxx(vec) *max_element(vec.begin(),vec.end())
using namespace std;
#ifdef LOCAL
template<typename T> void _print(T x) { cerr << x; }
// 可以打印多层数组，但是格式不是很好
template<typename T> void _print(vector<T> v) { cerr << "[ "; for (T i : v) _print(i), cerr << " "; cerr << "]"; }
template<typename T> void _print(set<T> s) { cerr << "{ "; for (T i : s) _print(i), cerr << " "; cerr << "}"; }
template<typename T, typename U> void _print(pair<T, U> p) { cerr << "{"; _print(p.first); cerr << ", "; _print(p.second); cerr << "}"; }
template<typename T, typename U> void _print(map<T, U> m) { cerr << "{ "; for (auto &p : m) _print(p), cerr << " "; cerr << "}"; }
// 打印两层数组，格式好一些
template<typename T> void _print(vector<vector<T>> v){cerr<<"{\n";for (auto i : v) _print(i), cerr << "\n";cerr<<" }";}
#endif

using  ll = long long;using ull = unsigned long long;
using ld=double;using LD=long double;
using i128 = __int128;
using pdd = pair<ld,ld>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;using vb=vector<bool>;
using tll = tuple<ll,ll,ll>;using vii=vector<int>;
using vpll = vector<pll>;using vtll = vector<tll>;
using vdb = vector<ld>;using vc=vector<char>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
using CD=complex<double>;
static constexpr ll MAXN=(ll)1e6+10,INF=(ll)1e17+3; // 1e17+3是素数 1e18+9是素数
static constexpr ll mod=(ll)1e9+7; // 998244353;
static constexpr double eps=1e-8;const double pi=acos(-1.0);
ll N,M,Q,X,K,T,lT;

inline void __();
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    
    // cin>>lT;
    // while(lT--)
        __();
    return 0;
}
/*

*/
using Real = ld;
int sgn(Real x) {
    if (-eps < x && x < eps) return 0;
    return x < 0 ? -1 : 1;
}
struct Point{
    Real x,y;
    Point(){}
    Point(Real x,Real y):x(x),y(y){}
    //                      & 是指只对左值起效果，不能对Point(1,2)这样的临时值
    // friend Point& operator+=(Point &a,Point b){
    //     a.x+=b.x;
    //     a.y+=b.y;
    //     return a;
    // }
    friend Point operator +(Point a,Point b){
        return Point(a.x+b.x,a.y+b.y);
    }
    friend Point operator -(Point a,Point b){
        return Point(a.x-b.x,a.y-b.y);
    }
    friend bool operator <(Point a,Point b) {
        return sgn(a.x-b.x)?a.x<b.x:a.y<b.y; 
    }
    friend bool operator ==(Point a,Point b){
        if(sgn(a.x-b.x)) return false;
        if(sgn(a.y-b.y)) return false;
        return true;
    }
    friend Point operator *(Point a,Real k){
        return Point(a.x*k,a.y*k);
    }
    friend Point operator /(Point a,Real k){
        return Point(a.x/k,a.y/k);
    }
    friend bool operator !=(Point a,Point b){
        return !(a==b);
    }
    friend istream& operator >>(istream& is,Point &e){
        Real a,b;is>>a>>b;
        e=Point(a,b);
        return is;
    }
    friend ostream &operator<<(ostream &os, Point p) {
        return os << "(" << p.x << ", " << p.y << ")";
    }
};

Real cross(Point a,Point b){
    return a.x*b.y-a.y*b.x;
}
Real dot(Point a,Point b){
    return a.x*b.x+a.y*b.y;
}
// 采用两点式
struct Line{
    Point p,q;
    Line(){}
    Line(Point a,Point b):p(a),q(b) {}
    friend bool operator<(Line a,Line b){
        if(a.p!=b.p) return a.p<b.p;
        return a.q<b.q; 
    }
    friend bool operator==(Line a,Line b){
        if(a.p!=b.p) return false;
        if(a.q!=b.q) return false;
        return true;
    } 
    friend istream& operator>>(istream& is, Line& e) {
        Real a,b,c,d;
        is >> a >> b >> c >> d;
        e = Line(Point(a,b),Point(c,d));
        return is;
    }
    friend ostream&operator<<(ostream &os, Line l) {
        return os << "<" << l.p << ", " << l.q << ">";
    }
};

// p+tv=q+sw
// (p-q) x w+tv x w = 0
// t=-(p-q)x w/(vxw)
// 传入的是方向式  拿到的是两直线交点
Point getintersect(const Line &a,const Line &b){
    Point p=a.p, v=a.q-a.p,q=b.p,w=b.q-b.p;
    Point u=p-q;
    ld t=cross(w,u)/cross(v,w);
    return p+v*t;
}
bool isseginterline(Line seg,Line e){
    if(sgn(cross(e.p-seg.p, e.q-seg.p))*sgn(cross(e.p-seg.q, e.q-seg.q))>0){
        return false;
    }
    return true;
}
bool onseg(Point p,Line s){
    return sgn(cross(p-s.p, p-s.q))==0 && sgn(dot(p-s.p,p-s.q))<=0;
}
struct Dis{
    ll dis,idx;
    friend bool operator <(Dis a,Dis b){
        return a.dis>b.dis;
    }
};
inline void __(){
    cin>>N;
    vector<Line> A(N);
    vector<Point> node;
    vector<Point> b1,b2,b3,b4;
    auto add=[&](Point a){
        if(sgn(a.x)==0){
            b1.EB(a);
        }else if(sgn(a.x-100)==0){
            b2.EB(a);
        }else if(sgn(a.y)==0){
            b3.EB(a);
        }else{
            b4.EB(a);
        }
    };
    FOR(i,0,N-1){
        cin>>A[i];
        add(A[i].p);
        add(A[i].q);
    }
    FOR(i,0,N-1){
        vector<Point> wp;
        Line a=A[i];
        FOR(j,0,N-1){
            if(i==j) continue;
            Line b=A[j];
            if(isseginterline(a, b) && isseginterline(b, a)){
                wp.PB(getintersect(a, b));
            }
        }
        wp.EB(a.p);wp.EB(a.q);
        sort(all(wp));
        wp.resize(unique(all(wp))-wp.begin());
        FOR(i,1,SZ(wp)-1){
            Point mid=(wp[i]+wp[i-1])/2;
            node.EB(mid);
        }
    }
    sort(all(node));node.resize(unique(all(node))-node.begin());
    Point sta;cin>>sta;node.EB(sta);
    ll st=SZ(node);
    // node.EB(Point(0,0));
    // node.EB(Point(0,100));
    // node.EB(Point(100,100));
    // node.EB(Point(100,0));
    auto addnode=[&](vector<Point> & a){
        sort(all(a));
        a.resize(unique(all(a))-a.begin());
        FOR(i,1,SZ(a)-1){
            Point mid=(a[i]+a[i-1])/2;
            node.EB(mid);
        }
    };
    b1.EB(0,0);b1.EB(0,100);
    addnode(b1);
    b2.EB(100,0);b2.EB(100,100); 
    addnode(b2);
    b3.EB(0,0);b3.EB(100,0);
    addnode(b3);
    b4.EB(0,100);b4.EB(100,100);
    addnode(b4);
    auto check=[&](Line a)->bool {
        bool ok=true;
        FOR(i,0,SZ(A)-1){
            Line t=A[i];
            // if(t==a) continue;
            if(onseg(a.p, t) && onseg(a.q, t)) return false;
            if(isseginterline(t, a) && isseginterline(a, t)){
                if(onseg(a.p, t) || onseg(a.q, t)) continue;
                ok=false;
                break;
            }
        }
        return ok;
    };
    vector<vector<int> > g(SZ(node)+6);
    FOR(i,0,SZ(node)-1){
        FOR(j,i+1,SZ(node)-1){
            if(check(Line(node[i],node[j]))){
                g[i].EB(j);
                g[j].EB(i);
            }
        }
    }
// #ifdef LOCAL
//     _print(g);
//     _print(b1);
//     debug(sta);
//     debug(st);
// #endif
    priority_queue<Dis> pq;
    pq.push({0,st-1});
    vector<ll> dist(SZ(node)+5,INF);
    while (!pq.empty()) {
        auto [dis,u]=pq.top();pq.pop();
        if(dist[u]!=INF) continue;
        dist[u]=dis;
        for(auto v:g[u]){
            #ifdef LOCAL
                debug(v);
            #endif
            pq.push({dis+1,v});
        }
    }
    ll ans=*min_element(dist.begin()+st,dist.end());
// #ifdef LOCAL
//     _print(dist);
// #endif
    cout<<"Number of doors = "<<ans<<"\n";
}
/*
AC
https://vjudge.net/solution/63304444

*/

https://vjudge.net/problem/UVA-754

这道题目的UVA版本好像还加了数据，继续测一下。

搞了半天是格式问题，要换两次行。

https://vjudge.net/solution/63313110

源代码

// by Gospel_rock
#include <algorithm>
#include <bits/stdc++.h>
#include <cstdio>
#include <istream>
#include <ostream>
#include <queue>
#include <vector>
#define all(vec) vec.begin(),vec.end()
#define PB push_back
#define EB emplace_back
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((long long) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define fsp(x) fixed<<setprecision(x)
#define minn(vec) *min_element(vec.begin(),vec.end())
#define maxx(vec) *max_element(vec.begin(),vec.end())
using namespace std;
#ifdef LOCAL
template<typename T> void _print(T x) { cerr << x; }
// 可以打印多层数组，但是格式不是很好
template<typename T> void _print(vector<T> v) { cerr << "[ "; for (T i : v) _print(i), cerr << " "; cerr << "]"; }
template<typename T> void _print(set<T> s) { cerr << "{ "; for (T i : s) _print(i), cerr << " "; cerr << "}"; }
template<typename T, typename U> void _print(pair<T, U> p) { cerr << "{"; _print(p.first); cerr << ", "; _print(p.second); cerr << "}"; }
template<typename T, typename U> void _print(map<T, U> m) { cerr << "{ "; for (auto &p : m) _print(p), cerr << " "; cerr << "}"; }
// 打印两层数组，格式好一些
template<typename T> void _print(vector<vector<T>> v){cerr<<"{\n";for (auto i : v) _print(i), cerr << "\n";cerr<<" }";}
#endif

using  ll = long long;using ull = unsigned long long;
using ld=double;using LD=long double;
using i128 = __int128;
using pdd = pair<ld,ld>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;using vb=vector<bool>;
using tll = tuple<ll,ll,ll>;using vii=vector<int>;
using vpll = vector<pll>;using vtll = vector<tll>;
using vdb = vector<ld>;using vc=vector<char>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
using CD=complex<double>;
static constexpr ll MAXN=(ll)1e6+10,INF=(ll)1e17+3; // 1e17+3是素数 1e18+9是素数
static constexpr ll mod=(ll)1e9+7; // 998244353;
static constexpr double eps=1e-9;const double pi=acos(-1.0);
ll N,M,Q,X,K,T,lT;

inline void __();
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    
    cin>>lT;
    while(lT--)
        __();
    return 0;
}
/*

*/
using Real = ld;
int sgn(Real x) {
    if (-eps < x && x < eps) return 0;
    return x < 0 ? -1 : 1;
}
struct Point{
    Real x,y;
    Point(){}
    Point(Real x,Real y):x(x),y(y){}
    //                      & 是指只对左值起效果，不能对Point(1,2)这样的临时值
    // friend Point& operator+=(Point &a,Point b){
    //     a.x+=b.x;
    //     a.y+=b.y;
    //     return a;
    // }
    friend Point operator +(Point a,Point b){
        return Point(a.x+b.x,a.y+b.y);
    }
    friend Point operator -(Point a,Point b){
        return Point(a.x-b.x,a.y-b.y);
    }
    friend bool operator <(Point a,Point b) {
        return sgn(a.x-b.x)?a.x<b.x:a.y<b.y; 
    }
    friend bool operator ==(Point a,Point b){
        if(sgn(a.x-b.x)) return false;
        if(sgn(a.y-b.y)) return false;
        return true;
    }
    friend Point operator *(Point a,Real k){
        return Point(a.x*k,a.y*k);
    }
    friend Point operator /(Point a,Real k){
        return Point(a.x/k,a.y/k);
    }
    friend bool operator !=(Point a,Point b){
        return !(a==b);
    }
    friend istream& operator >>(istream& is,Point &e){
        Real a,b;is>>a>>b;
        e=Point(a,b);
        return is;
    }
    friend ostream &operator<<(ostream &os, Point p) {
        return os << "(" << p.x << ", " << p.y << ")";
    }
};

Real cross(Point a,Point b){
    return a.x*b.y-a.y*b.x;
}
Real dot(Point a,Point b){
    return a.x*b.x+a.y*b.y;
}
// 采用两点式
struct Line{
    Point p,q;
    Line(){}
    Line(Point a,Point b):p(a),q(b) {}
    friend bool operator<(Line a,Line b){
        if(a.p!=b.p) return a.p<b.p;
        return a.q<b.q; 
    }
    friend bool operator==(Line a,Line b){
        if(a.p!=b.p) return false;
        if(a.q!=b.q) return false;
        return true;
    } 
    friend istream& operator>>(istream& is, Line& e) {
        Real a,b,c,d;
        is >> a >> b >> c >> d;
        e = Line(Point(a,b),Point(c,d));
        return is;
    }
    friend ostream&operator<<(ostream &os, Line l) {
        return os << "<" << l.p << ", " << l.q << ">";
    }
};

// p+tv=q+sw
// (p-q) x w+tv x w = 0
// t=-(p-q)x w/(vxw)
// 传入的是方向式  拿到的是两直线交点
Point getintersect(const Line &a,const Line &b){
    Point p=a.p, v=a.q-a.p,q=b.p,w=b.q-b.p;
    Point u=p-q;
    ld t=cross(w,u)/cross(v,w);
    return p+v*t;
}
bool seginterline(Line seg,Line e){
    if(sgn(cross(e.p-seg.p, e.q-seg.p))*sgn(cross(e.p-seg.q, e.q-seg.q))>0){
        return false;
    }
    return true;
}
bool seginterseg(Line a,Line b){
    // auto [s,e]=a;
    // auto [s2,e2]=b;
    // Real A=max(s.x,e.x),AA=min(s2.x,e2.x);
    // Real B=max(s.y,e.y),BB=min(s2.y,e2.y);
    // if(sgn(A-AA)<0 && sgn(B-BB)<0){
    //     return false;
    // }
    if(seginterline(a, b) && seginterline(b, a)){
        return true;
    }
    return false;
}
bool onseg(Point p,Line s){
    return sgn(cross(p-s.p, p-s.q))==0 && sgn(dot(p-s.p,p-s.q))<=0;
}
struct Dis{
    ll dis,idx;
    friend bool operator <(Dis a,Dis b){
        return a.dis>b.dis;
    }
};
inline void __(){
    cin>>N;
    vector<Line> A(N);
    vector<Point> node;
    vector<Point> b1,b2,b3,b4;
    auto add=[&](Point a){
        if(sgn(a.x)==0){
            b1.EB(a);
        }else if(sgn(a.x-100)==0){
            b2.EB(a);
        }else if(sgn(a.y)==0){
            b3.EB(a);
        }else{
            b4.EB(a);
        }
    };
    FOR(i,0,N-1){
        cin>>A[i];
        add(A[i].p);
        add(A[i].q);
    }
    FOR(i,0,N-1){
        vector<Point> wp;
        Line a=A[i];
        FOR(j,0,N-1){
            if(i==j) continue;
            Line b=A[j];
            if(sgn(cross(a.p-a.q,b.p-b.q))==0) continue;
            if(seginterseg(a, b)){
                wp.PB(getintersect(a, b));
            }
        }
        wp.EB(a.p);wp.EB(a.q);
        sort(all(wp));
        wp.resize(unique(all(wp))-wp.begin());
        FOR(i,1,SZ(wp)-1){
            Point mid=(wp[i]+wp[i-1])/2;
            node.EB(mid);
        }
    }
    sort(all(node));node.resize(unique(all(node))-node.begin());
    Point sta;cin>>sta;node.EB(sta);
    ll st=SZ(node);
    // node.EB(Point(0,0));
    // node.EB(Point(0,100));
    // node.EB(Point(100,100));
    // node.EB(Point(100,0));
    auto addnode=[&](vector<Point> & a){
        sort(all(a));
        a.resize(unique(all(a))-a.begin());
        FOR(i,1,SZ(a)-1){
            Point mid=(a[i]+a[i-1])/2;
            node.EB(mid);
        }
    };
    b1.EB(0,0);b1.EB(0,100);
    addnode(b1);
    b2.EB(100,0);b2.EB(100,100); 
    addnode(b2);
    b3.EB(0,0);b3.EB(100,0);
    addnode(b3);
    b4.EB(0,100);b4.EB(100,100);
    addnode(b4);
    auto check=[&](Line a)->bool {
        bool ok=true;
        FOR(i,0,SZ(A)-1){
            Line t=A[i];
            // if(t==a) continue;
            if(onseg(a.p, t) && onseg(a.q, t)) return false;
            if(seginterseg(a, t)){
                if(onseg(a.p, t) || onseg(a.q, t)) continue;
                ok=false;
                break;
            }
        }
        return ok;
    };
    vector<vector<int> > g(SZ(node)+6);
    FOR(i,0,SZ(node)-1){
        FOR(j,i+1,SZ(node)-1){
            if(check(Line(node[i],node[j]))){
                g[i].EB(j);
                g[j].EB(i);
            }
        }
    }
// #ifdef LOCAL
//     _print(g);
//     _print(b1);
//     debug(sta);
//     debug(st);
// #endif
    priority_queue<Dis> pq;
    pq.push({0,st-1});
    vector<ll> dist(SZ(node)+5,INF); 
    while (!pq.empty()) {
        auto [dis,u]=pq.top();pq.pop();
        if(dist[u]!=INF) continue;
        dist[u]=dis;
        for(auto v:g[u]){
            pq.push({dis+1,v});
        }
    }
    ll ans=*min_element(dist.begin()+st,dist.end());
// #ifdef LOCAL
//     _print(dist);
// #endif
    cout<<"Number of doors = "<<ans<<"\n";
    if(lT!=0) cout<<"\n";
}
/*
https://vjudge.net/problem/UVA-754



2
0 100 50 0
100 100 50 0
50 99

*/

心路历程（WA，TLE，MLE……）

Pick-up sticks

Posted on 2025-08-28 Edited on 2026-02-18

思路讲解

可以假设顶层木棍不超过1000根。

这句话比较重要。于是我们选择去维护这个顶层木棍，不断用上面的木棍去看这些木棍是不是顶层木棍。

核心代码长这样。

FOR(i,0,N-1){
    Line a=A[i];
    for(auto it=ans.begin();it!=ans.end();){
        Line t=A[*it];
        auto nit=next(it);
        if(isseginterline(a,t) && isseginterline(t,a)){
            ans.erase(it);
        }
        it=nit;
    }
    ans.EB(i);
}

AC代码

https://vjudge.net/solution/63275995

源代码

// by Gospel_rock
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define PB push_back
#define EB emplace_back
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((long long) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define fsp(x) fixed<<setprecision(x)
#define minn(vec) *min_element(vec.begin(),vec.end())
#define maxx(vec) *max_element(vec.begin(),vec.end())
using namespace std;
#ifdef LOCAL
template<typename T> void _print(T x) { cerr << x; }
// 可以打印多层数组，但是格式不是很好
template<typename T> void _print(vector<T> v) { cerr << "[ "; for (T i : v) _print(i), cerr << " "; cerr << "]"; }
template<typename T> void _print(set<T> s) { cerr << "{ "; for (T i : s) _print(i), cerr << " "; cerr << "}"; }
template<typename T, typename U> void _print(pair<T, U> p) { cerr << "{"; _print(p.first); cerr << ", "; _print(p.second); cerr << "}"; }
template<typename T, typename U> void _print(map<T, U> m) { cerr << "{ "; for (auto &p : m) _print(p), cerr << " "; cerr << "}"; }
// 打印两层数组，格式好一些
template<typename T> void _print(vector<vector<T>> v){cerr<<"{\n";for (auto i : v) _print(i), cerr << "\n";cerr<<" }";}
#endif

using  ll = long long;using ull = unsigned long long;
using ld=double;using LD=long double;
using i128 = __int128;
using pdd = pair<ld,ld>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;using vb=vector<bool>;
using tll = tuple<ll,ll,ll>;using vii=vector<int>;
using vpll = vector<pll>;using vtll = vector<tll>;
using vdb = vector<ld>;using vc=vector<char>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
using CD=complex<double>;
static constexpr ll MAXN=(ll)1e6+10,INF=(ll)1e17+3; // 1e17+3是素数 1e18+9是素数
static constexpr ll mod=(ll)1e9+7; // 998244353;
static constexpr double eps=1e-8;const double pi=acos(-1.0);
ll N,M,Q,X,K,T,lT;

inline void __();
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    
    // cin>>lT;
    while(cin>>N)
        __();
    return 0;
}
/*

*/
using Real = ld;
using Point = complex<Real>;
#define x() real()
#define y() imag()
Real cross(const Point &a, const Point &b) {
    return (conj(a) * b).imag();
} 
Real dot(const Point &a, const Point &b) {
    return (conj(a) * b).real();
}
int sgn(Real x) {
    if (-eps < x && x < eps) return 0;
    return x < 0 ? -1 : 1;
}
namespace std
{
    bool operator <(const Point& a, const Point& b){
        return sgn(real(a) - real(b)) ? real(a) < real(b) : imag(a) <
               imag(b);
    }
    bool operator ==(const Point& a, const Point& b){
        if(!sgn(a.imag()-b.imag()) && !sgn(a.real()-b.real())){
            return true;
        }
        return false;
    }
    bool operator !=(const Point& a, const Point& b){
        if(!sgn(a.imag()-b.imag()) && !sgn(a.real()-b.real())){
            return false;
        }
        return true;
    }
    istream& operator>>(istream& in, Point& c) {
        Real r, i;
        in >> r >> i;
        c = Point(r, i);
        return in;
    }
}
// 采用两点式
struct Line{
    Point a,b;
    Line(){}
    Line(Point a,Point b):a(a),b(b) {}
    bool operator<(const Line&other) const{
        return a!=other.a?a<other.a:b<other.b;
    }
    bool operator==(const Line&other) const{
        return a==other.a && b==other.b;
    }
};
istream& operator>>(istream& in, Line& line) {
    Real a,b,c,d;
    in >> a >> b >> c >> d;
    line = Line(Point(a,b),Point(c,d));
    return in;
}
// p+tv=q+sw
// (p-q) x w+tv x w = 0
// t=-(p-q)x w/(vxw)
// 传入的是方向式  拿到的是两直线交点
Point getintersect(const Line &a,const Line &b){
    Point p=a.a, v=a.b-a.a,q=b.a,w=b.b-b.a;
    Point u=p-q;
    ld t=cross(w,u)/cross(v,w);
    return p+v*t;
}
bool isseginterline(Line seg,Line line){
    if(sgn(cross(line.a-seg.a, line.b-seg.a))*sgn(cross(line.a-seg.b, line.b-seg.b))>0){
        return false;
    }
    return true;
}
inline void __(){
    if(N==0) return;
    vector<Line> A;
    FOR(i,1,N){
        Line a;
        cin>>a;
        A.EB(a);
    }
    vb vis(N+5);
    list<int> ans;
    cout<<"Top sticks: ";
    // ROF(i,SZ(ans)-1,0){
    //     if(i!=0)cout<<ans[i]<<", ";
    //     else cout<<ans[i]<<".\n";
    // }
    FOR(i,0,N-1){
        Line a=A[i];
        for(auto it=ans.begin();it!=ans.end();){
            Line t=A[*it];
            auto nit=next(it);
            if(isseginterline(a,t) && isseginterline(t,a)){
                ans.erase(it);
            }
            it=nit;
        }
        ans.EB(i);
    }
    for(auto it=ans.begin();it!=ans.end();++it){
        if(it!=(--ans.end())){
            cout<< *it+1<<", ";
        }else{
            cout<< *it+1<<".\n";
        }
    }

}
/*
AC
https://vjudge.net/solution/63275995
*/

心路历程（WA，TLE，MLE……）

The Doors

Posted on 2025-08-28 Edited on 2026-02-18

思路讲解

可以用dij做。搜索做不太了。

先建边，然后再跑dij，会快一些。

auto check=[&](Line a){
    int idx=lower_bound(all(A),Line(a.a,Point(INF,INF)))-A.begin();
    bool ok=true;
    FOR(i,idx,SZ(A)-1){
        Line t=A[i];
        if(sgn(t.a.x()-a.b.x())>=0) break;
        if(isseginterline(t, a)){
            ok=false;
            break;
        }
    }
    return ok;
};
vb vis(node.size()+5);
vector<vector<pair<ld,int> > > g(SZ(node)+5);
priority_queue<DIS> pq;
pq.EM(0,0);
FOR(i,0,SZ(node)-1){
    FOR(j,i+1,SZ(node)-1){
        if(node[i].x()==node[j].x()){
            continue;
        }
        Point a=node[i],b=node[j];
        if(check(Line(a,b))){
            g[i].emplace_back(abs(a-b),j);
        }
    }
}

AC代码

https://vjudge.net/solution/63271204

源代码

// by Gospel_rock
#include <algorithm>
#include <bits/stdc++.h>
#include <queue>
#include <utility>
#include <vector>
#define all(vec) vec.begin(),vec.end()
#define PB push_back
#define EB emplace_back
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((long long) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define fsp(x) fixed<<setprecision(x)
#define minn(vec) *min_element(vec.begin(),vec.end())
#define maxx(vec) *max_element(vec.begin(),vec.end())
using namespace std;
#ifdef LOCAL
template<typename T> void _print(T x) { cerr << x; }
// 可以打印多层数组，但是格式不是很好
template<typename T> void _print(vector<T> v) { cerr << "[ "; for (T i : v) _print(i), cerr << " "; cerr << "]"; }
template<typename T> void _print(set<T> s) { cerr << "{ "; for (T i : s) _print(i), cerr << " "; cerr << "}"; }
template<typename T, typename U> void _print(pair<T, U> p) { cerr << "{"; _print(p.first); cerr << ", "; _print(p.second); cerr << "}"; }
template<typename T, typename U> void _print(map<T, U> m) { cerr << "{ "; for (auto &p : m) _print(p), cerr << " "; cerr << "}"; }
// 打印两层数组，格式好一些
template<typename T> void _print(vector<vector<T>> v){cerr<<"{\n";for (auto i : v) _print(i), cerr << "\n";cerr<<" }";}
#endif

using  ll = long long;using ull = unsigned long long;
using ld=double;using LD=long double;
using i128 = __int128;
using pdd = pair<ld,ld>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;using vb=vector<bool>;
using tll = tuple<ll,ll,ll>;using vii=vector<int>;
using vpll = vector<pll>;using vtll = vector<tll>;
using vdb = vector<ld>;using vc=vector<char>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
using CD=complex<double>;
static constexpr ll MAXN=(ll)1e6+10,INF=(ll)1e17+3; // 1e17+3是素数 1e18+9是素数
static constexpr ll mod=(ll)1e9+7; // 998244353;
static constexpr double eps=1e-8;const double pi=acos(-1.0);
ll N,M,Q,X,K,T,lT;

inline void __();
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    
    // cin>>lT;
    while(cin>>N)
        __();
    return 0;
}
/*

*/
using Real = ld;
using Point = complex<Real>;
#define x() real()
#define y() imag()
Real cross(const Point &a, const Point &b) {
    return (conj(a) * b).imag();
} 
Real dot(const Point &a, const Point &b) {
    return (conj(a) * b).real();
}
int sgn(Real x) {
    if (-eps < x && x < eps) return 0;
    return x < 0 ? -1 : 1;
}
namespace std
{
    bool operator <(const Point& a, const Point& b){
        return sgn(real(a) - real(b)) ? real(a) < real(b) : imag(a) <
               imag(b);
    }
    bool operator ==(const Point& a, const Point& b){
        if(!sgn(a.imag()-b.imag()) && !sgn(a.real()-b.real())){
            return true;
        }
        return false;
    }
    bool operator !=(const Point& a, const Point& b){
        if(!sgn(a.imag()-b.imag()) && !sgn(a.real()-b.real())){
            return false;
        }
        return true;
    }
}
istream& operator>>(istream& in, Point& c) {
    double r, i;
    in >> r >> i;
    c = Point(r, i);
    return in;
}
// 采用两点式
struct Line{
    Point a,b;
    Line(){}
    Line(Point a,Point b):a(a),b(b) {}
    bool operator<(const Line&other) const{
        return a!=other.a?a<other.a:b<other.b;
    }
    bool operator==(const Line&other) const{
        return a==other.a && b==other.b;
    }
};
// p+tv=q+sw
// (p-q) x w+tv x w = 0
// t=-(p-q)x w/(vxw)
// 传入的是方向式  拿到的是两直线交点
Point getintersect(const Line &a,const Line &b){
    Point p=a.a, v=a.b-a.a,q=b.a,w=b.b-b.a;
    Point u=p-q;
    ld t=cross(w,u)/cross(v,w);
    return p+v*t;
}
bool isseginterline(Line seg,Line line){
    if(sgn(cross(line.a-seg.a, line.b-seg.a))*sgn(cross(line.a-seg.b, line.b-seg.b))>0){
        return false;
    }
    return true;
}
struct DIS{
    ld dis;
    int idx;
    bool operator<(const DIS &other) const{
        return dis>other.dis;
    }
    DIS(ld dis,int idx):dis(dis),idx(idx) {}
};
inline void __(){
    if(N==-1) return;
    vector<Line> A;
    vector<Point> node={Point(0,5)};
    FOR(i,1,N){
        ld x,y1,y2,y3,y4;
        cin>>x>>y1>>y2>>y3>>y4;
        Point a(x,0),b(x,y1);
        A.emplace_back(a,b);
        node.emplace_back(b);
        Point a2(x,y2),b2(x,y3);
        A.emplace_back(a2,b2);
        node.emplace_back(a2);
        node.emplace_back(b2);
        Point a3(x,y4),b3(x,10);
        A.emplace_back(a3,b3);
        node.emplace_back(a3);
    }
    ld ans=INF;
    Point dest(10,5);
    node.EB(dest);
    auto check=[&](Line a){
        int idx=lower_bound(all(A),Line(a.a,Point(INF,INF)))-A.begin();
        bool ok=true;
        FOR(i,idx,SZ(A)-1){
            Line t=A[i];
            if(sgn(t.a.x()-a.b.x())>=0) break;
            if(isseginterline(t, a)){
                ok=false;
                break;
            }
        }
        return ok;
    };
    vb vis(node.size()+5);
    vector<vector<pair<ld,int> > > g(SZ(node)+5);
    priority_queue<DIS> pq;
    pq.EM(0,0);
    FOR(i,0,SZ(node)-1){
        FOR(j,i+1,SZ(node)-1){
            if(node[i].x()==node[j].x()){
                continue;
            }
            Point a=node[i],b=node[j];
            if(check(Line(a,b))){
                g[i].emplace_back(abs(a-b),j);
                
#ifdef LOCAL
    debug(j);debug(a);debug(b);
#endif
            }
        }
    }
    vector<ld> Dis(SZ(node)+5); 
    while (!pq.empty()) {
        auto [dis,ind]=pq.top();pq.pop();
#ifdef LOCAL
    debug(dis);
    debug(ind);
#endif
        // if(vis.count(a)) continue;
        // vis.insert(a);
        Point a=node[ind];
        if(vis[ind]){
            continue;
        }
        vis[ind]=true;
        Dis[ind]=dis;
        for(auto [w,v]:g[ind]){
// #ifdef LOCAL
//     debug(w);debug(v);
// #endif
            pq.EM(dis+w,v);
        }
    }
    cout<<fsp(2)<<Dis[SZ(node)-1]<<"\n";
#ifdef LOCAL
    cend;
#endif
}
/*
1
2
4 2 7 8 9
7 3 4.5 6 7
-1
AC
https://vjudge.net/solution/63271204
*/

心路历程（WA，TLE，MLE……）

Intersecting Lines（两线是否相交）

Posted on 2025-08-27 Edited on 2026-02-18

思路讲解

详细解释了一下这个式子是怎么来的。

// p+tv=q+sw
// (p-q) x w+tv x w = 0
// t=-(p-q)x w/(vxw)
// 传入的是方向式
Point getintersect(Point p,Point v,Point q,Point w){
    Point u=p-q;
    ld t=cross(w,u)/cross(v,w);
    return p+v*t;
}
inline void __(){
    Point a,b,c,d;
    cin>>a>>b>>c>>d;
    Point ab=b-a,cd=d-c;
    if(!sgn(cross(ab,cd))){
        Point bc=c-b,ad=d-a;
        if(!sgn(cross(bc, ad))){
            cout<<"LINE\n";
            return;
        }
        cout<<"NONE\n";
        return;
    }
    Point res=getintersect(a, ab, c, cd);
    cout<<"POINT "<<fsp(2)<<res.real()<<" "<<res.imag()<<"\n";
}

AC代码

https://vjudge.net/solution/63256517

源代码

// by Gospel_rock
#include <bits/stdc++.h>
#include <cstdio>
#define all(vec) vec.begin(),vec.end()
#define PB push_back
#define EB emplace_back
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((long long) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define fsp(x) fixed<<setprecision(x)
#define minn(vec) *min_element(vec.begin(),vec.end())
#define maxx(vec) *max_element(vec.begin(),vec.end())
using namespace std;
#ifdef LOCAL
template<typename T> void _print(T x) { cerr << x; }
// 可以打印多层数组，但是格式不是很好
template<typename T> void _print(vector<T> v) { cerr << "[ "; for (T i : v) _print(i), cerr << " "; cerr << "]"; }
template<typename T> void _print(set<T> s) { cerr << "{ "; for (T i : s) _print(i), cerr << " "; cerr << "}"; }
template<typename T, typename U> void _print(pair<T, U> p) { cerr << "{"; _print(p.first); cerr << ", "; _print(p.second); cerr << "}"; }
template<typename T, typename U> void _print(map<T, U> m) { cerr << "{ "; for (auto &p : m) _print(p), cerr << " "; cerr << "}"; }
// 打印两层数组，格式好一些
template<typename T> void _print(vector<vector<T>> v){cerr<<"{\n";for (auto i : v) _print(i), cerr << "\n";cerr<<" }";}
#endif

using  ll = long long;using ull = unsigned long long;
using ld=double;using LD=long double;
using i128 = __int128;
using pdd = pair<ld,ld>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;using vb=vector<bool>;
using tll = tuple<ll,ll,ll>;using vii=vector<int>;
using vpll = vector<pll>;using vtll = vector<tll>;
using vdb = vector<ld>;using vc=vector<char>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
using CD=complex<double>;
static constexpr ll MAXN=(ll)1e6+10,INF=(ll)1e17+3; // 1e17+3是素数 1e18+9是素数
static constexpr ll mod=(ll)1e9+7; // 998244353;
static constexpr double eps=1e-8;const double pi=acos(-1.0);
ll N,M,Q,X,K,T,lT;

inline void __();
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    
    cin>>lT;
    cout<<"INTERSECTING LINES OUTPUT\n";
    while(lT--)
        __();
    cout<<"END OF OUTPUT\n";
    return 0;
}
/*

*/
using Real = ld;
using Point = complex<Real>;
Real cross(const Point &a, const Point &b) {
    return (conj(a) * b).imag();
} 
Real dot(const Point &a, const Point &b) {
    return (conj(a) * b).real();
}
int sgn(Real x) {
    if (-eps < x && x < eps) return 0;
    return x < 0 ? -1 : 1;
}
istream& operator>>(istream& in, Point& c) {
    double r, i;
    in >> r >> i;
    c = Point(r, i);
    return in;
}
// p+tv=q+sw
// (p-q) x w+tv x w = 0
// t=-(p-q)x w/(vxw)
// 传入的是方向式
Point getintersect(Point p,Point v,Point q,Point w){
    Point u=p-q;
    ld t=cross(w,u)/cross(v,w);
    return p+v*t;
}
inline void __(){
    Point a,b,c,d;
    cin>>a>>b>>c>>d;
    Point ab=b-a,cd=d-c;
    if(!sgn(cross(ab,cd))){
        Point bc=c-b,ad=d-a;
        if(!sgn(cross(bc, ad))){
            cout<<"LINE\n";
            return;
        }
        cout<<"NONE\n";
        return;
    }
    Point res=getintersect(a, ab, c, cd);
    cout<<"POINT "<<fsp(2)<<res.real()<<" "<<res.imag()<<"\n";
}
/*

*/