hdoj 5533 Dancing Stars on Me

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Dancing Stars on Me

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 586 Accepted Submission(s): 310

Problem Description

The sky was brushed clean by the wind and the stars were cold in a black sky. What a wonderful night. You observed that, sometimes the stars can form a regular polygon in the sky if we connect them properly. You want to record these moments by your smart camera. Of course, you cannot stay awake all night for capturing. So you decide to write a program running on the smart camera to check whether the stars can form a regular polygon and capture these moments automatically.

Formally, a regular polygon is a convex polygon whose angles are all equal and all its sides have the same length. The area of a regular polygon must be nonzero. We say the stars can form a regular polygon if they are exactly the vertices of some regular polygon. To simplify the problem, we project the sky to a two-dimensional plane here, and you just need to check whether the stars can form a regular polygon in this plane.

Input

The first line contains a integer

T indicating the total number of test cases. Each test case begins with an integer

n, denoting the number of stars in the sky. Following

n lines, each contains

2 integers

xi,yi, describe the coordinates of

n stars.

1≤T≤300

3≤n≤100

−10000≤xi,yi≤10000
All coordinates are distinct.

Output

For each test case, please output "`YES`" if the stars can form a regular polygon. Otherwise, output "`NO`" (both without quotes).

Sample Input

330 01 11 040 00 11 01 150 00 10 22 22 0

Sample Output

NOYESNO
主要是方法问题：
      for（i=0;i<n;i++）
        for(j=i+1;j<n;j++)
   将每个点都与没有用过的进行求距离，求得的距离排序（从小到大），看最小的数有多少个！（即最小且相等的边的个数）如果==输入的点数
  证明是正n边行！
     代码 ：
             #include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>using namespace std; struct node{double  x;double  y;}num[1100]; //double  dist(int  i,int  j)//不要调用函数，不然超时！ //{//return  sqrt((num[i].x -num[j].x )*(num[i].x -num[j].x )+(num[i].y -num[j].y )*(num[i].y -num[j].y ));//}int main(){int  T;scanf("%d",&T);while(T--)  {int  n,i,j;scanf("%d",&n);for(i=0;i<n;i++){scanf("%lf%lf",&num[i].x ,&num[i].y );}double  b[11000];int  k=0;for(i=0;i<n;i++)  for(j=i+1;j<n;j++)  {   b[k++]=sqrt((num[i].x -num[j].x )*(num[i].x -num[j].x )+(num[i].y -num[j].y )*(num[i].y -num[j].y ));  }  sort(b,b+k);  int m=1;  for(i=1;i<n;i++)  {  if(b[i]==b[i-1])  m++;  else  break;  }  if(m==n)  printf("YES\n");  else  printf("NO\n");  }  return  0;}

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