nyoj 1099 Lan Xiang's Square （给出四个点的坐标，检测能否组成正方形）

实验室小学妹这次腾讯模拟笔试中遇到相似的题，于是找了一道，拿来做一下。。。。

Lan Xiang’s Square
时间限制：1000 ms | 内存限制：65535 KB
难度：0

描述

Excavator technology which is strong, fast to Shandong to find Lan Xiang.

Then the question comes.. :)

for this problem , i will give you four points. you just judge if they can form a square.
if they can, print “Yes”, else print “No”.
Easy ? just AC it.

输入
T <= 105 cases.
for every case
four points, and every point is a grid point .-10^8 <= all interger <= 10^8。
grid point is both x and y are interger.
输出
Yes or No
样例输入

1
1 1
-1 1
-1 -1
1 -1

样例输出

Yes

提示
you think this is a easy problem ? you dare submit, i promise you get a WA. :)

给出四个点的坐标，检测能否组成正方形

思路：先排序，再证明四个边相等，再证明两个对角线相等。。关键点，一定要先排序，先排序，先排序。。。。

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
struct Point{
    double x,y;
    Point(double _x, double _y):x(_x),y(_y){};
    bool operator < (const Point &a) const{
        return x==a.x?y<a.y:x<a.x;
    }
};
int main()
{
    int n;
    cin>>n;
    while(n--)
    {
        vector<Point> a;
        double xx, yy;
        for (int i = 0; i < 4; ++i) {
            cin>>xx>>yy;
            a.push_back(Point(xx, yy));
        }
        sort(a.begin(), a.end());
        double l1, l2, l3, l4, l5, l6;
        l1=(a[0].x-a[2].x)*(a[0].x-a[2].x)+(a[0].y-a[2].y)*(a[0].y-a[2].y);
        l2=(a[0].x-a[1].x)*(a[0].x-a[1].x)+(a[0].y-a[1].y)*(a[0].y-a[1].y);
        l3=(a[3].x-a[1].x)*(a[3].x-a[1].x)+(a[3].y-a[1].y)*(a[3].y-a[1].y);
        l4=(a[2].x-a[3].x)*(a[2].x-a[3].x)+(a[2].y-a[3].y)*(a[2].y-a[3].y);
        l5 = (a[1].x-a[2].x)*(a[1].x-a[2].x)+(a[1].y-a[2].y)*(a[1].y-a[2].y);
        l6 = (a[0].x-a[3].x)*(a[0].x-a[3].x)+(a[0].y-a[3].y)*(a[0].y-a[3].y);
        if(l1==l2&&l3==l4&&l1==l3&&l1!=0&&l5==l6)
            cout<<"Yes"<<endl;
        else
            cout<<"No"<<endl;
    }
}