HDU-5563 (Clarke and five-pointed star)

Problem Description
Clarke is a patient with multiple personality disorder. One day, Clarke turned into a learner of geometric.
When he did a research with polygons, he found he has to judge if the polygon is a five-pointed star at many times. There are 5 points on a plane, he wants to know if a five-pointed star existed with 5 points given.


Input
The first line contains an integer T(1≤T≤10), the number of the test cases.
For each test case, 5 lines follow. Each line contains 2 real numbers xi,yi(−109≤xi,yi≤109), denoting the coordinate of this point.


Output
Two numbers are equal if and only if the difference between them is less than 10−4.
For each test case, print Yes if they can compose a five-pointed star. Otherwise, print No. (If 5 points are the same, print Yes. )


Sample Input
2
3.0000000 0.0000000
0.9270509 2.8531695
0.9270509 -2.8531695
-2.4270509 1.7633557
-2.4270509 -1.7633557
3.0000000 1.0000000
0.9270509 2.8531695
0.9270509 -2.8531695
-2.4270509 1.7633557

-2.4270509 -1.7633557

题意：给出任意顺序的5个点的坐标，判断5个点是否能连接成五角星，误差小于10^-4;

思路：首先想到的是判断是否为正五边形，想到用向量来解，结果因为坐标不是按照顺序给出的，，于是放弃这个思路。。。

然后从边的角度考虑，即判断边是否相等，要判断的边共有10条，里边五角星5条，外边围成的正五边形5条，，

若是五角星，则前后5条边必定不相等，且5条边内部分别相等，于是先排序，

再分别判断前5条边之间与后5条边之间是否相等。。即误差小于10^ -4;

以下AC代码：

#include<stdio.h>#include<stdlib.h>#include<math.h>int cmp(const void *a,const void *b){    return *(double *)a > *(double *)b ? 1 : -1;}int main(){    int t;    double x[5],y[5];    double a[15];    int i,j;    scanf("%d",&t);    while(t--)    {      for(i=0;i<5;i++)      {          scanf("%lf%lf",&x[i],&y[i]);      }      int temp=0;      int flag=1;       for(i=0;i<5;i++)        for(j=0;j<5,i!=j;j++)       {           a[temp++]=sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));   //求出10条边       }       qsort(a,temp,sizeof(a[0]),cmp);        //从小到大排序        for(i=0;i<4;i++)        {                                                                  //判断前5条边之间是否相等            if(fabs(a[i]-a[i+1])>0.0001)                {                  flag=0;                  break;                }        }        for(i=5;i<9;i++)        {                                                               //判断后5条边之间是否相等            if(fabs(a[i]-a[i+1])>0.0001)            {                flag=0;                break;            }        }        if(flag)            printf("Yes\n");        else            printf("No\n");    }    return 0;}