Dancing Stars on Me HDU - 5533

最新推荐文章于 2018-09-15 11:09:15 发布

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本文介绍了一种利用智能相机自动捕捉天空中星星形成的正多边形图案的方法。通过编程实现对星星坐标数据的处理，判断星星是否能构成正多边形，并自动记录这些特殊时刻。

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 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

Sample Output

NO
YES
NO

Think:判断是否能够组成正多边形，只要最短的边的个数等于坐标个数就成立，反之不成立

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
struct node{
    int x,y;
}a[110];
int b[110][110];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        for (int i=0;i<n;i++){
            scanf("%d%d",&a[i].x,&a[i].y);
        }
        int x=800000000;
        for (int i=0;i<n;i++){
            for (int j=i+1;j<n;j++){
                b[i][j]=(a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y);
                if (x>b[i][j]){
                    x=b[i][j];
                }
            }
        }
        int count=0;
        for (int i=0;i<n;i++){
            for (int j=i+1;j<n;j++){
                if (b[i][j]==x){
                    count++;
                }
            }
        }
        if (count==n){
            printf("YES\n");
        }
        else{
            printf("NO\n");
        }
    }
    return 0;
}