Znzryb's Blog

Pick-up sticks

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

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 2026-04-06

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 2026-04-06

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";
}
/*

*/

心路历程（WA，TLE，MLE……）

Segments

Posted on 2025-08-27 Edited on 2026-04-06

思路讲解

样例一就是选一条垂线就行。

通过样例一不难发现，只要存在一条线与所有的线段相交，那么，就可以了。

因为所有与这条线段相交的点都会投影到这条线的垂线上。

AC代码

https://vjudge.net/solution/63255893

源代码

// by Gospel_rock
#include <bits/stdc++.h>
#include <cstdlib>
#include <utility>
#include <vector>
#define all(vec) vec.begin(),vec.end()
#define PB push_back
#define EB emplace_back
#define EM emplace
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((long long) a.size())
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define FI first
#define SE second
#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;
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;
}
// int dcmp(double x, double y){  //比较两个浮点数：0 相等；-1 小于；1 大于
//     if(fabs(x - y) < eps) return 0;
//     else return x<y ?-1:1;
// }


inline void __(){
    cin>>N;
    vector<pair<Point, Point> > A;
    vector<Point> ls;
    for(int i=1;i<=N;++i){
        ld a,b,c,d;
        cin>>a>>b>>c>>d;
        Point x(a,b),y(c,d);
        A.emplace_back(x,y);
// #ifdef LOCAL
//     debug(x);debug(y);
// #endif
        ls.EB(x);ls.EB(y);
    }
    for(int i=0;i<SZ(ls);++i){
        for(int j=i+1;j<SZ(ls);++j){
            bool ok=true;
            Point a=ls[i],b=ls[j];
            // 不能是同一点
            if(!sgn(a.real()-b.real()) && !sgn(a.imag()-b.imag())) continue;
            for(int k=0;k<N;++k){
                Point x=A[k].FI,y=A[k].SE;
                if(sgn(cross(a-x,b-x))*sgn(cross(a-y,b-y))>0){
                    ok=false;
                    break;
                }
            }
            if(ok){
                cout<<"Yes!\n";
                return;
            }
        }
    }
    cout<<"No!\n";
}
/*
AC
https://vjudge.net/solution/63255893
*/


//     for(int i=0;i<N;++i){
//         ld x=A[i].FI.real();
//         Point ab=A[i].SE-A[i].FI;
//         Point ab1=ab/abs(ab);
//         vector<pair<ld, ld>> wp;
//         for(int j=0;j<N;++j){
//             ld l=dot(A[j].first-A[i].first, ab1)*ab1.real()+x;
//             ld r=dot(A[j].second-A[i].first, ab1)*ab1.real();
// #ifdef LOCAL
//     debug(l);debug(r);
// #endif
//             wp.emplace_back(l,r);
//         }
//         ld l=wp[0].FI,r=wp[0].SE;
//         bool ok=true;
//         for(int j=1;j<N;++j){
//             ld le=wp[j].FI,rr=wp[j].SE;
//             if(sgn(rr-l)<0 || sgn(le-r)>0){
//                 ok=false;
//                 break;
//             }
//             l=max(l,le),r=min(r,rr);
//         }
//         if(ok){
//             cout<<"Yes!\n";
//             return;
//         }
//     }

心路历程（WA，TLE，MLE……）

P11217 【MX-S4-T1】「yyOI R2」youyou 的垃圾桶

Posted on 2025-08-26 Edited on 2026-04-06

思路讲解

线段树上二分

https://www.luogu.com.cn/article/2a6kvgjo

参考这篇题解。

当然，我们先要枚举一下需要操作几轮。

FOR(j,1,64){
    if(w>sum){
        w-=sum;
        ans+=N;
    }else{
        break;
    }
    bei*=2;
    sum*=2;
}

AC代码

https://www.luogu.com.cn/record/233323384

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 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 =  __int128;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,W;

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

*/
ll bei=1;
// 参考zkw的PPT 以及 https://www.luogu.com.cn/record/207546670
// 参考题解 https://www.luogu.com.cn/article/w3gfeebh
// AC https://www.luogu.com.cn/record/232827448 区间加
constexpr ll preset=0;
struct SEG{
    static ll op(const ll &a,const ll &b){
        return a+b;
    }
    int n,sz=1,dep;
    vector<ll> tr;vector<ll> tag;
    SEG(int n){     // 传入的时候可以传大一点
        this->n=n+2;
        dep=__lg(this->n)+1;
        sz=1<<dep;
        tr.resize((sz<<1)+5,preset);
        tag.resize((sz<<1)+5,preset);
    }
    void pushup(int o){tr[o]=op(tr[o<<1],tr[o<<1|1]);}
    void settag(int o,ll x){
        tr[o]=op(tr[o],x);
        tag[o]=op(tag[o],x);
    }
    void pushdown(int o){
        int l=o<<1,r=l|1;   // 左儿子，右儿子
        settag(l, tag[o]/2);
        settag(r, tag[o]/2);
        tag[o]=0;
    }
    void assign(int i,ll x){
        tr[i+sz]=x;
    }
    void build(){
        for(int o=sz-1;o>=1;--o){
            pushup(o);
        }
    }
    void push(int l,int r){
        //        在j层，l，r变为了同一点
        int i=dep,j=__lg(l^r);
        while (i>j) {   // 就是为了避免一下重复下推
            pushdown(l>>i--);
        }
        while(i){
            pushdown(l>>i),pushdown(r>>i--);
        }
    }
    ll query(int l,int r){
        ll res=preset;
        // 变为开区间，并加上sz
        l=l-1+sz,r=r+1+sz;
        push(l,r);
        for(;l^r^1;l>>=1,r>>=1){
            // l^r^1 表示 只要 l 与 r 还不是同一个父亲的两兄弟，就继续上跳。
            if(~l&1){   
                // l是偶数，说明l是父节点的左儿子，开区间，将其右儿子加入答案
                res=op(res,tr[l^1]);
            }
            if(r&1){
                res=op(tr[r^1],res);
            }
        }
        return res;
    }
    void add(int l,int r,ll x){
        int w=1;
        l=l-1+sz,r=r+1+sz;
        push(l, r);
        for(;l^r^1;pushup(l>>=1),pushup(r>>=1),w<<=1){
            if(~l&1){
                settag(l^1, w*x);
            }
            if(r&1){
                settag(r^1, w*x);
            }
        }
        for (l >>= 1; l; l >>= 1) pushup(l);
    }
    // AC https://qoj.ac/submission/1263556
    // 找到最靠近r的不符合位置 fun return true 是不符合
    template<class F> int findl(int r,F&&fun){
        return findl(1,r,0,sz-1,0,fun);
    }
    // 找到最靠近r的不符合位置
    template<class F> int findl(int o,int qr,int l,int r,ll sum,F&&fun){
        if(l==r){
            return l;
        }
        pushdown(o);
        ll ls=o<<1,rs=ls|1;
        ll mid=l+r>>1;
        ll pos=-1;
        if(fun(tr[rs]*bei+sum+tr[ls]*bei) && mid+1<=qr) 
            pos=findl(rs,qr,mid+1,r,sum+tr[ls]*bei,fun);
        if(pos!=-1) return pos;
        if(fun(tr[ls]*bei+sum)) 
            pos=findl(ls,qr,l,mid,sum,fun);
        return pos;
    }
    // 找最靠近l的不符合位置
    template<class F> int findr(int l,F&&fun){
        return findr(1,l,0,sz-1,0,fun);
    }
    // 找最靠近l的不符合位置
    template<class F> int findr(int o,int ql,int l,int r,ll sum,F&&fun){
        if(l==r){
            return l;
        }
        pushdown(o);
        ll ls=o<<1,rs=ls|1;
        ll mid=l+r>>1;
        ll pos=-1;
        if(fun(tr[ls]*bei+sum) && mid>=ql) pos=findr(ls,ql,l,mid,sum,fun);
        if(pos!=-1) return pos;
        if(fun(tr[rs]*bei+sum+tr[ls]*bei)) pos=findr(rs,ql,mid+1,r,sum+tr[ls]*bei,fun);
        return pos;
    }
};
// 来自wida github仓库 https://github.com/hh2048/XCPC/tree/main 加了一些新功能
//46 ms https://acm.hdu.edu.cn/contest/view-code?cid=1179&rid=14912 读入字符getc(a) 的AC记录
// AC https://vjudge.net/solution/63197697 一道测试读写的codechef题目
namespace QuickRead { // 快读
    char buf[1 << 21], *p1 = buf, *p2 = buf;
    inline int getc() {
        return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
    }
    template <typename T> void Cin(T &a) {
        T ans = 0;
        bool f = 0;
        char c = getc();
        for (; c < '0' || c > '9'; c = getc()) {
            if (c == '-') f = 1;
        }
        for (; c >= '0' && c <= '9'; c = getc()) {
            ans = ans * 10 + c - '0';
        }
        a = f ? -ans : ans;
    }
    void getc(char &a){
        char c=getc();
        while (isspace(c)) {
            c=getc();
        }
        a=c;
    }
    template <typename T, typename... Args> void Cin(T &a, Args &...args) {
        Cin(a), Cin(args...);
    }
    // ===== 输出 =====
    static constexpr size_t OUT_CAP = 1 << 21;  // 大概是2MB
    char obuf[OUT_CAP];
    size_t op = 0;
    inline void flush() {
        if (op) {
            fwrite(obuf, 1, op, stdout);
            op = 0;
        }
    }
    inline void putc(char c) {
        if (op == OUT_CAP) flush();
        obuf[op++] = c;
    }
    template <typename T> void write(T x) { // 注意，这里输出不带换行
        if (x < 0) putc('-'), x = -x;
        if (x > 9) write(x / 10);
        putc(x % 10 + '0');
    }
    struct Flusher {    // 析构时自动刷新
        ~Flusher() { flush(); }
    } flusher;
} // namespace QuickRead
using namespace QuickRead;

inline void __(){
    Cin(N,Q,W);
    SEG tr(N+1);
    FOR(i,1,N){
        ll a;Cin(a);
        tr.assign(i,a);
    }
    // tr.assign(N+1, INF);
    tr.build();
    FOR(i,1,Q){
        ll l,r,d;
        Cin(l,r,d);
        tr.add(l, r, d);
        long long ans=0;
        ll sum=tr.tr[1];
        ll w=W;
        bei=1;
        FOR(j,1,64){
            if(w>sum){
                w-=sum;
                ans+=N;
            }else{
                break;
            }
            bei*=2;
            sum*=2;
        }
#ifdef LOCAL
    debug(ans);
    debug( (long long)w );
    debug((long long)tr.tr[1])
#endif
        ans+=tr.findr(1,[&](ll x){
            if(x>=w) return true;
            return false;
        })-1;
#ifdef LOCAL
    debug(ans);
    cend;
#endif
        write(ans);putc('\n');
    }
}
/*
AC
https://www.luogu.com.cn/record/233323384
*/