NYOJ 1099 Lan Xiang's Square（给出四点判断是否能构成正方形）



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. :)

来源

myself

上传者

ACM_张开创

对于给的四个点首先应该先进行排序，排序时自己一定要清楚排的顺序，不然会影响后面的计算的。然后就是分别把四条边以及两条对角线的长度给求出来。下面开始对边进行判断，四条边都相等且两条对角线相等（当然长度不会为0，这个要考虑到）的图形可以构成正方形。

代码：

#include<stdio.h>#include<algorithm>#include<math.h>#include<string.h>using namespace std;struct A{double x,y;}a[4];bool cmp (A a,A b){if(a.x != b.x)return a.x < b.x;elsereturn a.y > b.y; }int main(){int t;scanf("%d",&t);while(t--){for(int i = 0;i < 4;i++)scanf("%lf%lf",&a[i].x,&a[i].y);sort(a,a + 4,cmp);//四个点先排序 //for(int i = 0;i < 4;i++)//printf(" ##  %lf   %lf\n",a[i].x,a[i].y);double l1,l2,l3,l4,l5,l6; //l1~l4为四条边，l5.l6为两条对角线的长度 l1 = sqrt(pow((a[0].x - a[1].x),2) + pow((a[0].y - a[1].y),2));l2 = sqrt(pow((a[0].x - a[2].x),2) + pow((a[0].y - a[2].y),2));l3 = sqrt(pow((a[2].x - a[3].x),2) + pow((a[2].y - a[3].y),2));l4 = sqrt(pow((a[1].x - a[3].x),2) + pow((a[1].y - a[3].y),2));l5 = sqrt(pow((a[0].x - a[3].x),2) + pow((a[0].y - a[3].y),2));l6 = sqrt(pow((a[1].x - a[2].x),2) + pow((a[1].y - a[2].y),2));if(l5 != l6 || l5 == 0) //两对角线之间的判断 {printf("No\n");}else if(l5 == l6)  //对角线相等的情况下判断4条边 {if(l1 == l2 && l1 == l3 && l1 == l4)printf("Yes\n");elseprintf("No\n");}}return 0;}