HDU5533 Dancing Stars on Me(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): 1800    Accepted Submission(s): 1045

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

 

题意：给你n颗星星的坐标求星星是否能构成正n边形

思路：因为是n个星星就要正n边形，而且给你的点的坐标都是整数，所以只有当n为4时才可能构成正n边形，即正方形，其他形状时必有两边不相等而不得其他边形，所以当n为4时将两两个点的距离计算出来然后排个序，若为正方形，则四边相等对角线相等

#include <iostream>#include <cstdio>#include <cstring>#include <stdio.h>#include <stdlib.h>#include <algorithm>#include <math.h>using namespace std;struct point{    double x,y;};double dis(point a,point b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n;        scanf("%d",&n);        point a[n];        for(int i=0;i<n;i++)            scanf("%lf%lf",&a[i].x,&a[i].y);        if(n!=4)        {            printf("NO\n");            continue;        }        double b[6];        b[0]=dis(a[0],a[1]);        b[1]=dis(a[1],a[2]);        b[2]=dis(a[2],a[3]);        b[3]=dis(a[0],a[2]);        b[4]=dis(a[0],a[3]);        b[5]=dis(a[1],a[3]);        sort(b,b+6);        if(b[0]==b[1]&&b[1]==b[2]&&b[2]==b[3]&&b[4]==b[5])            printf("YES\n");        else            printf("NO\n");    }    return 0;}