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  • 稳定凸包——poj1228

    严格凸包的情况下,每条边上至少要有三个点,叫做稳定凸包

    #include<iostream>
    #include<cstring>
    #include<cstdio>
    #include<cmath>
    #include<vector>
    #include<algorithm>
    using namespace std;
    #define N 3005
    typedef double db;
    const db eps=1e-6;
    const db pi=acos(-1);
    int sign(db k){
        if (k>eps) return 1; else if (k<-eps) return -1; return 0;
    }
    int cmp(db k1,db k2){return sign(k1-k2);}
    int inmid(db k1,db k2,db k3){return sign(k1-k3)*sign(k2-k3)<=0;}// k3 在 [k1,k2] 内 
    struct point{
        db x,y;
        point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
        point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
        point operator * (db k1) const{return (point){x*k1,y*k1};}
        point operator / (db k1) const{return (point){x/k1,y/k1};}
        bool operator < (const point k1) const{
            int a=cmp(x,k1.x);
            if (a==-1) return 1; else if (a==1) return 0; else return cmp(y,k1.y)==-1;
        }
        db abs(){return sqrt(x*x+y*y);}
        db abs2(){return x*x+y*y;}
        db dis(point k1){return ((*this)-k1).abs();}
    };
    int inmid(point k1,point k2,point k3){return inmid(k1.x,k2.x,k3.x)&&inmid(k1.y,k2.y,k3.y);}
    db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}
    db dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;} 
    
    int onS(point k1,point k2,point q){return inmid(k1,k2,q)&&sign(cross(k1-q,k2-k1))==0;}
    int checkconvex(vector<point>A){ //判断是否是凸包
        int n=A.size(); A.push_back(A[0]); A.push_back(A[1]);
        for (int i=0;i<n;i++) if (sign(cross(A[i+1]-A[i],A[i+2]-A[i]))==-1) return 0;
        return 1;
    }
    vector<point> ConvexHull(vector<point>A,int flag=1){ // 求凸包:flag=0 不严格 flag=1 严格 
        int n=A.size(); vector<point>ans(n*2); 
        sort(A.begin(),A.end()); int now=-1;
        for (int i=0;i<A.size();i++){
            while (now>0&&sign(cross(ans[now]-ans[now-1],A[i]-ans[now-1]))<flag) now--;
            ans[++now]=A[i];
        } int pre=now;
        for (int i=n-2;i>=0;i--){
            while (now>pre&&sign(cross(ans[now]-ans[now-1],A[i]-ans[now-1]))<flag) now--;
            ans[++now]=A[i];
        } ans.resize(now); return ans;
    }
    
    vector<point> v;
    int n;
    
    int main(){
        int t;cin>>t;
        while(t--){
            cin>>n;
            v.clear();
            for(int i=1;i<=n;i++){
                point pp;
                scanf("%lf%lf",&pp.x,&pp.y);
                v.push_back(pp);
            }
            
            if(n<=4){puts("NO");continue;}
            
            vector<point> res=ConvexHull(v,0);
            if(res.size()!=v.size()){
                puts("NO");continue;
            }
            res.push_back(res[0]);
            res.push_back(res[1]);
            res.push_back(res[2]);
            
            int Max=0,len=0;//记录连续转角 
            for(int i=0;i<res.size()-2;i++){
                if(sign(cross(res[i+1]-res[i],res[i+2]-res[i]))!=0){
                    len++;Max=max(Max,len);
                }
                else len=0;
            }
            if(Max<=1)puts("YES");
            else puts("NO");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zsben991126/p/12361209.html
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