poj 1228 求一个稳定凸包

最新推荐文章于 2021-02-17 19:02:18 发布

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本文介绍了一个算法问题，即如何通过剩余的钉子来确定农场边界的独特性。输入包括钉子的数量及坐标，输出则是边界是否能被唯一确定。

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

Sample Output

NO

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <math.h>

using namespace std;

const int N = 40005;

typedef double DIY;

struct Point
{
    DIY x,y;
};

Point p[N];
Point stack[N];
Point MinA;

int top;

DIY dist(Point A,Point B)
{
    return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y));
}

DIY cross(Point A,Point B,Point C)
{
    return (B.x-A.x)*(C.y-A.y)-(B.y-A.y)*(C.x-A.x);
}

bool cmp(Point a,Point b)
{
    DIY k=cross(MinA,a,b);
    if(k>0) return 1;
    if(k<0) return 0;
    return dist(MinA,a)<dist(MinA,b);  //这里共线的点按距离从小到大排序
}

void Graham(int n)
{
    int i;
    for(i=1; i<n; i++)
        if(p[i].y<p[0].y||(p[i].y==p[0].y&&p[i].x<p[0].x))
            swap(p[i],p[0]);
    MinA=p[0];
    sort(p+1,p+n,cmp);
    stack[0]=p[0];
    stack[1]=p[1];
    top=1;
    for(i=2; i<n; i++)
    {
        //注意这里我们把共线的点也压入凸包里
        while(cross(stack[top-1],stack[top],p[i])<0&&top>=1) --top;
        stack[++top]=p[i];
    }
}

bool Judge()
{
    for(int i=1;i<top;i++)
    {
        //判断凸包的一条边上是否至少有3点
        if((cross(stack[i-1],stack[i+1],stack[i]))!=0&&(cross(stack[i],stack[i+2],stack[i+1]))!=0)
            return false;
    }
    return true;
}

int main()
{
    int t,n,i;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=0;i<n;i++)
            scanf("%lf%lf",&p[i].x,&p[i].y);
        if(n<6)
        {
            puts("NO");
            continue;
        }
        Graham(n);
        /*cout<<endl;
        for(i=0;i<n;i++)
          cout<<p[i].x<<" "<<p[i].y<<endl;
        cout<<endl;
        for(i=0;i<=top;i++)
          cout<<stack[i].x<<" "<<stack[i].y<<endl;
          */
        if(Judge()) puts("YES");
        else        puts("NO");
    }
    return 0;
}