NYOJ 1099 Lan Xiang's Square (判断是否为正四边形)

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. :)

水

大意：给出四个点，判断围成的图形是否为正四边形

思路：直接判断是否最短边存在4条，最长边存在两条就行了，可以自己手绘一下

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define MAXN 201000
#define MAX(a,b) a>b?a:b
#define fab(a) ((a)>0?(a):-(a))
#define mem(x) memset(x,0,sizeof(x))
#define INF 0xfffffff
using namespace std;
struct s
{
	double x,y;
}a[5];
double dis[MAXN];
double fun(s aa,s bb)
{
	return sqrt((aa.x-bb.x)*(aa.x-bb.x)+(aa.y-bb.y)*(aa.y-bb.y));
}
int main()
{
	int t,i,j;
	scanf("%d",&t);
	while(t--)
	{
		for(i=1;i<=4;i++)
		scanf("%lf%lf",&a[i].x,&a[i].y);
		double M=-1.0,mi=INF*1.0;
		int k=0;
		for(i=1;i<=4;i++)
		{
			for(j=i+1;j<=4;j++)
			{
				dis[k]=fun(a[i],a[j]);
				M=max(dis[k],M);
				mi=min(dis[k],mi);
				k++;
			}
		}
		int b=0,c=0;
		for(i=0;i<k;i++)
		{
			if(dis[i]==M)
			b++;
			else if(dis[i]==mi)
			c++;
		}
		if(b==2&&c==4)
		printf("Yes\n");
		else
		printf("No\n");
	}
	return 0;
}