Grandpa's Estate POJ

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稳定的凸包满足：在加新点使凸包扩大时，新凸包无法包含原来的所有顶点

换句话说，一个稳定的凸包的每一条边上都有至少三个顶点

题目数据比较水，没有所有点共线的特判（非稳定），我也就懒得写了

//#include<bits/stdc++.h>  //#pragma comment(linker, "/STACK:1024000000,1024000000")   #include<stdio.h>  #include<algorithm>  #include<queue>  #include<string.h>  #include<iostream>  #include<math.h>  #include<set>  #include<map>  #include<vector>  #include<iomanip>  using namespace std;    const double pi=acos(-1.0);  #define ll long long  #define pb push_back#define sqr(a) ((a)*(a))#define dis(a,b) sqrt(sqr(a.x-b.x)+sqr(a.y-b.y))const double eps=1e-6;const int maxn=1e3+56;const int inf=0x3f3f3f3f;int n;int tot;//凸包上点数struct Point{double x,y;Point(){}Point(double x,double y):x(x),y(y){}}point[maxn],vertex[maxn];bool cmp(Point a,Point b){return(a.y<b.y||(a.y== b.y && a.x<b.x));}double xmult(Point p1,Point p2,Point p3){//p3p1,p3p2的夹角测试return ( (p1.x-p3.x)*(p2.y-p3.y)-(p1.y-p3.y)*(p2.x-p3.x) );}//正表示p1在p2的顺时针方向int Andrew(){//返回凸包顶点数sort(point,point+n,cmp);int top=1;vertex[0]=point[0];vertex[1]=point[1];for(int i=2;i<n;i++){while(top && xmult(point[i],vertex[top],vertex[top-1])>eps)top--;vertex[++top]=point[i];}int len=top;vertex[++top]=point[n-2];for(int i=n-3;i>=0;i--){while(top!=len && xmult(point[i],vertex[top],vertex[top-1])>eps)top--;vertex[++top]=point[i];}return top;}bool judge(int n){//判凸包稳定for(int i=1;i<n;i++){if(fabs(xmult(vertex[i],vertex[i+1],vertex[i-1]))>eps&&   fabs(xmult(vertex[i],vertex[i+1],vertex[(i+2)%(n)]))>eps){return 0;}}return 1;}int main(){int T;scanf("%d",&T);while(T--){scanf("%d",&n);for(int i=0;i<n;i++){scanf("%lf%lf",&point[i].x,&point[i].y);}tot=Andrew();if(tot<6){printf("NO\n");continue;}if(judge(tot))printf("YES\n");else printf("NO\n");}}

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