POJ1228——Grandpa's Estate（计算几何，凸包）

Grandpa's Estate

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 12618 Accepted: 3540

Description

Being the only living descendant of his grandfather, Kamran the Believer inherited all of the grandpa's belongings. The most valuable one was a piece of convex polygon shaped farm in the grandpa's birth village. The farm was originally separated from the neighboring farms by a thick rope hooked to some spikes (big nails) placed on the boundary of the polygon. But, when Kamran went to visit his farm, he noticed that the rope and some spikes are missing. Your task is to write a program to help Kamran decide whether the boundary of his farm can be exactly determined only by the remaining spikes.

Input

The first line of the input file contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case contains an integer n (1 <= n <= 1000) which is the number of remaining spikes. Next, there are n lines, one line per spike, each containing a pair of integers which are x and y coordinates of the spike.

Output

There should be one output line per test case containing YES or NO depending on whether the boundary of the farm can be uniquely determined from the input.

Sample Input

16 0 01 23 42 02 4 5 0

Sample Output

NO

Source

Tehran 2002 Preliminary

给你一个凸包上面的点，在边界上或者在顶点上。询问这是不是一个稳定凸包。

构造一个凸包，判断一下每两个顶点之间是不是还有其他的点与之共线。所有点共线或者只有一个点的时候特判。

#include <stdio.h>#include <string.h>#include <vector>#include <queue>#include <algorithm>#include <cmath>using namespace std;#define MAXN 1010#define INF 1000000007#define EPS 1e-10//考虑误差的加减运算double add(double a,double b){    if(abs(a+b)<EPS*(abs(a)+abs(b)))        return 0;    else        return a+b;}struct P{    double x,y;    P(){}    P(double x,double y):x(x),y(y){}    P operator+(P p){        return P(add(x,p.x),add(y,p.y));    };    P operator-(P p){        return P(add(x,-p.x),add(y,-p.y));    };    P operator *(double d){        return P(x*d,y*d);    }    double dot(P p){ //求内积        return add(x*p.x,y*p.y);    }    double det(P p){ //求外积        return add(x*p.y,-y*p.x);    }};//字典序bool cmp_x(const P& p,const P&q){    if(p.x!=q.x) return p.x<q.x;    return p.y<q.y;}//求凸包vector <P> convex_hull(P* ps,int n){    sort(ps,ps+n,cmp_x);    int k=0; //凸包的顶点数    vector <P> qs(n*2); //构造中的凸包    //构造凸包的下侧    for(int i=0;i<n;i++){        while(k>1&&(qs[k-1]-qs[k-2]).det(ps[i]-qs[k-1])<0) k--;        qs[k++]=ps[i];    }    //构造凸包的上侧    for(int i=n-2,t=k;i>=0;i--){        while(k>t&&(qs[k-1]-qs[k-2]).det(ps[i]-qs[k-1])<0) k--;        qs[k++]=ps[i];    }    qs.resize(k-1);    return qs;}//输入int n;P ps[MAXN];bool solve(){    vector<P> qs=convex_hull(ps, n);    int m=(int)qs.size();    if(m==1)        return false;    int ok=0;    for(int i=0;i<m-2;i++)    {        if((qs[i+2]-qs[i+1]).det(qs[i+1]-qs[i])!=0){            ok=1;        }    }    if(ok==0)        return false;    for(int i=1;i<qs.size();i++){        if((qs[i]-qs[i-1]).det(qs[(i+1)%m]-qs[i])!=0&&((qs[(i+1)%m]-qs[i]).det(qs[(i+2)%m]-qs[(i+1)%m])!=0))            return false;    }    return true;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=0;i<n;i++)            scanf("%lf%lf",&ps[i].x,&ps[i].y);        if(solve())            printf("YES\n");        else            printf("NO\n");    }}