Grandpa's Estate

Grandpa's Estate

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 13640 Accepted: 3816

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

题意：给出一些点，判断凸包是否唯一

思路：判断凸包上的点中间是否有点，都有的话就稳定

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<queue>#include<stack>#include<vector>#include<cstring>#include<string>#include<algorithm>using namespace std;#define ll long long#define ms(a,b)memset(a,b,sizeof(a))#define eps 1e-10#define inf 1e8double a[4][4] = { {0,0,0,0},{0,-1,0,0},{0,0,0,-1},{0,-1,0,-1} } ;int n;double add(double a,double b){    if(abs(a+b)<eps*(abs(a)+abs(b))) return 0;    return a+b;}struct P{    double x,y;    double val;    double len;    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)    {        return add(x*p.x,y*p.y);    }    double det(P p)    {        return add(x*p.y,-y*p.x);    }}p[2000];bool on_seg(P p1,P p2,P q){    return (p1-q).det(p2-q)==0&&(p1-q).dot(p2-q)<0;}P intersection(P p1,P p2,P q1,P q2){    return p1+(p2-p1)*((q2-q1).det(q1-p1)/(q2-q1).det(p2-p1));}double ConvexPolygonArea(vector<P> p) {    double area = 0;    for(int i = 1; i+1 < p.size(); i++)        area+=(p[i]-p[0]).det(p[i + 1]-p[0]);    return area / 2;}bool cmp_x(const P& p,const P & q){    if(p.x!=q.x) return p.x<q.x;    return p.y<q.y;}vector<P> convex_hull(P *ps,int n){    sort(ps,ps+n,cmp_x);    int k=0;    vector<P> qs(n*2);    if(n==0) return qs;    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];    }    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;}int main(){    int n,t;    cin>>t;    while(t--)    {        scanf("%d",&n);        for(int i=0;i<n;i++)            scanf("%lf%lf",&p[i].x,&p[i].y);        vector<P> q=convex_hull(p,n);        int q_size=q.size();        if(q_size<3)        {            printf("NO\n");            continue;        }        bool flag=1;        for(int i=0;i<q_size;i++)        {            bool ok=false;            for(int j=0;j<n;j++)            {                if(on_seg(q[i],q[(i+1)%q_size],p[j]))                    ok=true;            }            if(!ok) flag=0;        }        if(flag) cout<<"YES"<<endl;        else cout<<"NO"<<endl;    }    return 0 ;}