Znzryb's Blog

P11361 [NOIP2024] 编辑字符串

Posted on 2025-09-02 Edited on 2026-04-06

思路讲解

要多写，如果实在没有思路就数学化。

参考思路如下。

https://www.luogu.com.cn/article/7xrsn748

不难发现，我们要最小化下面👇。

$\sum_{i=1}^{n}(s[i]-s'[i])^2 \Rightarrow \sum_{i=1}^{n}s[i]^2+s'[i]^2-2s[i]s'[i]$

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

常数分离以后，我们其实就是要最大化👇。

相当于我们要最大化 $\sum_{i=1}^{n}s[i]s'[i]$ 。不过我们发现好像没什么用？非常有用，我们就知道了，1 1 非常重要，要尽量多的出现1 1，其他不重要。

当然，如果光有这个想法是不行的。我们还需要用到分块思想。

首先那，分块，我们自然是同时只能按照一个字符串分块的，因此我们按照 $s$ 可不可以交换分块。

但是，我们按照 $s$ 分块以后，我们可以再对 $s'$ 进行分块。这个听起来很抽象是吧？其实也简单，就是说 $s’$ 其实如果不是落在整段 $s$ 可交换块上，可以部分落在 s 的可交换块上，然后我们的策略其实就是尽量多的出现1 1。

AC代码

源代码

心路历程（WA，TLE，MLE……）

至少有K个重复字符的最长子串

Posted on 2025-09-01 Edited on 2026-04-06

思路讲解

https://leetcode.cn/problems/longest-substring-with-at-least-k-repeating-characters/description/

其实就是双指针。

void cal(ll uni){
    vll cnt(28);
    ll l=0,r=0;
    ll cur=0,bigk=0;
    while (r<SZ(s)) {
        if(cur<=uni){
            int idx=s[r]-'a';
            if(cnt[idx]==0) ++cur;
            cnt[idx]++;
            if(cnt[idx]==K) ++bigk;
            ++r;
        }else{
            int idx=s[l]-'a';
            if(cnt[idx]==K) --bigk;
            --cnt[idx];
            if(cnt[idx]==0) --cur;
            ++l;
        }
        if(cur==uni && bigk==cur){
            ans=max(ans,r-l);
        }
    }
}

AC代码

https://leetcode.cn/problems/longest-substring-with-at-least-k-repeating-characters/submissions/658772401/

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 (long long i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (long long i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define lson(var) (var<<1)
#define rson(var) (var<<1|1)
#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 DB=double;using LD=long double;
// using i128 = __int128;
using pdd = pair<DB,DB>;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<DB>;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;
}
/*

*/
string s;
ll ans=0;
void cal(ll uni){
    vll cnt(28);
    ll l=0,r=0;
    ll cur=0,bigk=0;
    while (r<SZ(s)) {
        if(cur<=uni){
            int idx=s[r]-'a';
            if(cnt[idx]==0) ++cur;
            cnt[idx]++;
            if(cnt[idx]==K) ++bigk;
            ++r;
        }else{
            int idx=s[l]-'a';
            if(cnt[idx]==K) --bigk;
            --cnt[idx];
            if(cnt[idx]==0) --cur;
            ++l;
        }
        if(cur==uni && bigk==cur){
            ans=max(ans,r-l);
        }
    }
}
inline void __(){
    cin>>s;
    cin>>K;
    FOR(i,1,26){
        cal(i);
    }
    cout<<ans<<"\n";
}
/*

*/

心路历程（WA，TLE，MLE……）

ABC-421-D - RLE Moving

Posted on 2025-08-31 Edited on 2026-04-06

思路讲解

1
2

// 整体思路就是我们如果发现如果两者起始位置不同，那么是不会有相交的，或者相交一次
// 简单而言，就是必须先有一次相交，接着运动方向一样，才能有相交，否则就直接穿过去了

// 在同一点并且操作序列相同
if(a==b && opa==opb){
    ans+=mn;
}else{
    auto f=[&](ll dis){
        Point c=oper(a, opa,dis),d=oper(b, opb, dis);
        return mandis(c, d);
    };
    ll l=1,r=mn;
    while (l<r) {
        ll mid=l+r>>1;
        // if(f(mid)<=f(mid+1)){
        // 从mid点（或其前）开始，函数开始单调递增
        if(f(mid+1)-f(mid)>=0){
            r=mid;
        }else{
            l=mid+1;
        }
    }
    if(f(l)==0){
        ++ans;
    }
}

AC代码

https://atcoder.jp/contests/abc421/submissions/68978595

源代码

// by Gospel_rock
#include <bits/stdc++.h>
#include <deque>
#include <istream>
#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,L;

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

*/
using Real = ll;
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(b,a);
        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 cross(Point p1, Point p2, Point p0) { // 叉乘 (p1 - p0) x (p2 - p0);
    return cross(p1 - p0, p2 - p0);
}
Real dot(Point a,Point b){
    return a.x*b.x+a.y*b.y;
}
Real mandis(Point a,Point b){
    Real res=abs(a.x-b.x)+abs(a.y-b.y);
    return res;
}
// 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 onseg(Point p,Line s){
    return sgn(cross(p-s.p, p-s.q))==0 && sgn(dot(p-s.p,p-s.q))<=0;
}
// AC https://vjudge.net/problem/OpenJ_Bailian-1410
bool seginterseg(Line l1, Line l2) {
    Point s1 = l1.p, e1 = l1.q;
    Point s2 = l2.p, e2 = l2.q;
    // s2 相对于 s1->e1 其在哪一侧？
    int o1 = sgn(cross(e1, s2, s1));
    int o2 = sgn(cross(e1, e2, s1));
    int o3 = sgn(cross(e2, s1, s2));
    int o4 = sgn(cross(e2, e1, s2));
    // 防止把共线但不重叠的两线段判定为相交
    if (o1 == 0 && onseg(s2, l1)) return true;
    if (o2 == 0 && onseg(e2, l1)) return true;
    if (o3 == 0 && onseg(s1, l2)) return true;
    if (o4 == 0 && onseg(e1, l2)) return true;
    // 需要都在不同侧
    return (o1 * o2 < 0) && (o3 * o4 < 0);
}

Point oper(Point a,char op,ll dis){
    if(op=='U'){
        a.y-=dis;
    }else if(op=='D'){
        a.y+=dis;
    }else if(op=='R'){
        a.x+=dis;
    }else{
        a.x-=dis;
    }
    return a;
}
struct OP{
    char op;ll dis;
    friend istream& operator >>(istream &is,OP &e){
        is>>e.op>>e.dis;
        return is;
    }
};
inline void __(){
    Point a,b;
    cin>>a>>b;
    cin>>N>>M>>L;
    // 整体思路就是我们如果发现如果两者起始位置不同，那么是不会有相交的，或者相交一次
    // 简单而言，就是必须先有一次相交，接着运动方向一样，才能有相交，否则就直接穿过去了
    // vector<OP> A(N+5),B(N+5);
    queue<OP> A,B;
    FOR(i,1,M){
        OP a;
        cin>>a;
        A.push(a);
    }
    FOR(i,1,L){
        OP b;
        cin>>b;
        B.push(b);
    }
    ll ans=0;
    while (!A.empty()) {
        auto [opa,disa]=A.front();
        auto [opb,disb]=B.front();
        ll mn=min(disa,disb);
        A.front().dis-=mn;B.front().dis-=mn;
        if(A.front().dis==0){
            A.pop();
        }
        if(B.front().dis==0){
            B.pop();
        }
        if(a==b && opa==opb){
            ans+=mn;
        }else{
            auto f=[&](ll dis){
                Point c=oper(a, opa,dis),d=oper(b, opb, dis);
                return mandis(c, d);
            };
            ll l=1,r=mn;
            while (l<r) {
                ll mid=l+r>>1;
                // if(f(mid)<=f(mid+1)){
                // 从mid点（或其前）开始，函数开始单调递增
                if(f(mid+1)-f(mid)>=0){
                    r=mid;
                }else{
                    l=mid+1;
                }
            }
            if(f(l)==0){
                ++ans;
            }
        }
        a=oper(a, opa, mn);
        b=oper(b, opb, mn);
    }
    cout<<ans;
}
/*
AC
https://atcoder.jp/contests/abc421/submissions/68978595
*/

心路历程（WA，TLE，MLE……）

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

Posted on 2025-08-28 Edited on 2026-04-06

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 2026-04-06

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

*/