hdu5533Dancing Stars on Me(简单计算几何)

Dancing Stars on Me

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 599    Accepted Submission(s): 318

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 integern, 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

 题目大意：给出n边形的各顶点坐标，判断其是否可以组成一个正N边形；各坐标的值均为整数；

解题思路：由于个坐标值均为整数，只有正方形的所有坐标可以都为整数，所以就是判断能否构成正方形。

AC代码:

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct node{int x,y;}ans[120];int cmp(node a,node b){if(a.x!=b.x) return a.x<b.x;else return a.y<b.y;}int main(){int t;int n,i,j;scanf("%d",&t);while(t--){memset(ans,0,sizeof(ans));scanf("%d",&n);for(i=0;i<n;i++){scanf("%d%d",&ans[i].x,&ans[i].y);}   int s1,s2,s3,s4;   sort(ans,ans+n,cmp);   s1=ans[1].y-ans[0].y;   s2=ans[3].y-ans[2].y;   s3=ans[3].x-ans[1].x;   s4=ans[2].x-ans[0].x;   if(s1==s2&&s3==s4)    printf("YES\n");   else    printf("NO\n");}return 0;}