ZOJ2540 Form a Square

Form a Square

题意就是 判断 给你四个点，能否组成一个正方形

要点：

格式很重要， 很重要！！！

数据很小，直接暴力

四个点判断是否为正方形，只需将所有可能的边长度算出来，然后选其中最短的边作为正方形的边长，进行比较，看有没有符合的四条边

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
const int N = 4;
typedef struct aa{
    int x, y;
}AA;
AA a[N];

double dis(AA b, AA c)
{
    double len = sqrt((b.x - c.x) * (b.x - c.x) + (b.y - c.y) * (b.y - c.y));
    return len;
}
int main()
{
    int t, num, cnt = 1;
    cin >> t;
    int T = t;
    while(T--)
    {
        for(int i = 0; i < 4; i++)
        {
            cin >> a[i].x >> a[i].y;
        } 
        double ll = 2010;
        num = 0;
        for(int i = 0; i < 4; i++)
        {
            for(int j = i+1; j < 4; j++)
            {
                double dist = dis(a[i], a[j]);
                if(dist == ll)
                {
                    num++;
                }
                else if(dist < ll)
                {
                    ll = dist;
                    num = 1;
                }
            }
        }
        if(num == 4)
        {
            if(cnt != t)
                cout << "Case " << "" << cnt++ << ":" << endl << "Yes" << endl << endl;
            else
                cout << "Case " << "" << cnt++ << ":" << endl << "Yes";
        }
        else
        {
            if(cnt != t)
                cout << "Case " << "" << cnt++ << ":" << endl << "No" << endl << endl;
            else
                cout << "Case " << "" << cnt++ << ":" << endl << "No";
        }
        
    }
    return 0;
}