学步园

       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.

来源

myself

上传者

ACM_张开创

思路：被水题坑得好惨Orz 两领边相等
两对角线相等（ps：边可能为0，我就是死在这里 ） 点是乱序的 得先排序

（好像最先没有0这组数据
比赛时候我过了 后来重判把我给打回蓝翔了）

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<cmath>
#include<iostream>
using namespace std;

struct ac
{
    int  x,y;
} p[8];

bool cmp(ac xx,ac yy)
{
    if(xx.x!=yy.x)
        return xx.x<yy.x;
    return xx.y>yy.y;
}

int len(int a,int b,int c,int d)
{
    return sqrt((a-c)*(a-c)+(b-d)*(b-d));
}

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        for(int i=0; i<4; i++)
            cin>>p[i].x>>p[i].y;

        int s1,s2,s3,s4,s5,s6;
        sort(p,p+4,cmp);

        s1=len(p[0].x,p[0].y,p[3].x,p[3].y);
        s2=len(p[1].x,p[1].y,p[2].x,p[2].y);
        /*两邻边*/
        if(s1!=s2||s1==0)
        {
            cout<<"No"<<endl;
            continue;
        }
        else if(s1==s2)
        {
            if(((p[0].x-p[3].x)*(p[1].x-p[2].x)+(p[1].y-p[2].y)*(p[0].y-p[3].y))==0)//对角线相等
                cout<<"Yes"<<endl;
            else cout<<"No"<<endl;
        }
    }
}