P - Grandpa's Estate

 POJ - 1228 

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

1
6 
0 0
1 2
3 4
2 0
2 4 
5 0

Sample Output

NO

题意：给定一些点，判断这些点形成的凸包是否唯一

思路：如果凸包某条边上只有两个点，那么这些点形成的凸包不是唯一的， 因为丢掉的点可以是凸包外的，可以形成更大的凸包，但如果每条边都有3个以上点的话，不可能有丢掉的点在凸包外，假设有，那么该边中间的其他点就不在分界线上了，不符合实际。

#include<cstdio>
#include<stack>
#include<set>
#include<vector>
#include<queue>
#include<algorithm>
#include<cstring>
#include<string>
#include<map>
#include<iostream>
#include<cmath>
using namespace std;
#define inf 0x3f3f3f3f
typedef long long ll;
const int N=110;
const int nmax = 1020;
const double esp = 1e-9;
const double PI=3.1415926;
int n;
struct point
{
    int x,y;
    point()
    {
    }
    point(int _x,int _y):x(_x),y(_y)
    {
    }
    point operator -(const point &b)const
    {
        return point(x-b.x,y-b.y);
    }
} p[nmax],p0;
int tubao[nmax];
double cross(point p1,point p2)
{
    return p1.x*p2.y-p1.y*p2.x;
}
double dis(point p1,point p2)
{
    return sqrt((1.0*p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
bool cmp(point p1,point p2)
{
    double area=cross(p1-p0,p2-p0);
    if(fabs(area)<esp)
    {
        return dis(p1,p0)<dis(p2,p0);
    }
    if(area>esp)
        return true;
    return false;
}
void solve()
{
    int top=1;
    tubao[0]=0;
    tubao[1]=1;
    for(int i=2; i<n; i++)
    {
        while(top>0&&cross(p[tubao[top]]-p[tubao[top-1]],p[i]-p[tubao[top-1]])<0)
            //使凸包上同一条边3个以上点存在
            top--;
        tubao[++top]=i;
    }
    int flag=0;
    int s=0;
    tubao[++top]=0;
    for(int i=1; i<top; i++)
    {
        if(fabs(cross(p[tubao[i+1]]-p[tubao[i]],p[i]-p[tubao[s]]))<esp)  
        {
            flag=1;  //同一条线上多有两个以上的点
        }
        else if(flag==1)
        {
            s=i;
            flag=0;
        }
        else
        {
            flag=-1;
            break;
        }
    }
    if(flag==-1)
        printf("NO\n");
    else
        printf("YES\n");
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0; i<n; i++)
            scanf("%d%d",&p[i].x,&p[i].y);
        if(n<6)  //5个点以内不可能形成每条边有3个以上的点的凸包
        {
            printf("NO\n");
            continue;
        }
        int flag=0;
        for(int i=1; i<n-1; i++)
        {
            if(fabs(cross(p[i]-p[i-1],p[i+1]-p[i]))>esp)//判断是否可以形成凸包，即不能都同线
            {
                flag=1;  
                break;
            }
        }
        if(flag==0)
        {
            printf("NO\n");
            continue;
        }
        p0.y=inf;
        p0.x=0;
        int k=0;
        for(int i=1; i<n; i++)
        {
            if(p[i].y<p0.y||(p[i].y==p0.y&&p[i].x<p0.x))
            {
                p0=p[i];
                k=i;
            }
        }
        if(k)
        {
            point p1=p[k];
            p[k]=p[0];
            p[0]=p1;
        }
        sort(p+1,p+n,cmp);
        solve();
    }
    return 0;
}