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

11 1-1 1-1 -11 -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 0xfffffffusing 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");elseprintf("No\n");}return 0;}