• hdu 5533(几何水)


    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.

    1T300
    3n100
    10000xi,yi10000
    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
    3 3 0 0 1 1 1 0 4 0 0 0 1 1 0 1 1 5 0 0 0 1 0 2 2 2 2 0
     
    Sample Output
    NO YES NO



    题意:给你几个点,问能否形成正多边形。

    思路:

    ①找出所有点之间的最短距离,即边长。然后再在所有点中找出与边长相等的,只要数量为n即

    说明能够成正多边形

    ②要求输入的为整数,所以只有正四边形才能成立,判断下即可


    #include <iostream>
    #include <cstdio>
    #include <cmath>
    using namespace std;
    typedef long long ll;
    double x[1005];
    double y[1005];
    double len[1005][1005];
    
    double get_(int i,int j)
    {
        return sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
    }
    
    int main()
    {
        int T,n;
        scanf("%d",&T);
        while(T--)
        {
            double tmin = 0x3f3f3f3f;
            scanf("%d",&n);
            for(int i = 1; i <= n; i++)
            {
                scanf("%lf%lf",&x[i],&y[i]);
            }
            for(int i = 1; i <= n; i++)
                for(int j = i+1; j <= n; j++)
                {
                    len[i][j] = get_(i,j);
                    tmin = min(tmin,len[i][j]);
                }
            int num = 0;
            for(int i = 1; i <= n; i++)
                for(int j = i+1; j <= n; j++)
                {
                    if(tmin == len[i][j])
                    {
                        num++;
                    }
                }
            if(num == n)
                printf("YES
    ");
            else
                printf("NO
    ");
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/Przz/p/5409687.html
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