nyoj 1099 Lan Xiang's Square (水题)

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Lan Xiang's Square

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.

ACM_张开创

题意：给出四个点坐标，判断是否能组成正方形。

证明：四条边相等，并且有一个角是直角 为正方形。

题很简单 ，自己的思路也一直整错 可就是wa

可能数据太给力了，调试n次，发现sqrt()导致精度损失。 这个错误可真难找。。。

先贴上wa的代码

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#include 
      
       
#include 
       
         #define inf 500000005 int main() { double a[4],b[4],x1,x2,x3,x4,y1,y2,y3,y4; double sum1,sum2,l1,l2,l3,l4,l; int i,t; scanf("%d",&t); while(t--) { sum1=sum2=0,x1=x2=x3=x4=y1=y2=y3=y4=inf; for(i=0;i<4;i++) scanf("%lf %lf",&a[i],&b[i]),sum1+=a[i],sum2+=b[i]; sum1=sum1/4,sum2=sum2/4; for(i=0;i<4;i++) { if(a[i]
        
         =sum2) x1=a[i],y1=b[i]; if(a[i]<=sum1&&b[i]
         
          =sum1&&b[i]>sum2) x3=a[i],y3=b[i]; if(a[i]>sum1&&b[i]<=sum2) x4=a[i],y4=b[i]; } if(x1==inf||x2==inf||x3==inf||x4==inf||y1==inf||y2==inf||y3==inf||y4==inf) { printf("No\n"); continue; } l1=sqrt(pow(y1-y2,2)+pow(x1-x2,2)); l2=sqrt(pow(y2-y4,2)+pow(x2-x4,2)); l3=sqrt(pow(y3-y4,2)+pow(x3-x4,2)); l4=sqrt(pow(y1-y3,2)+pow(x1-x3,2)); l=sqrt(pow(y3-y2,2)+pow(x3-x2,2)); if(l1==l2&&l1==l3&&l1==l4&&fabs(l1*l1+l4*l4-l*l)<0.00001) printf("Yes\n"); else printf("No\n"); } return 0; } 
         
        
       
      

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#include 
      
       
#include 
       
         #define inf 500000005 int main() { double a[4],b[4],x1,x2,x3,x4,y1,y2,y3,y4; double sum1,sum2,l1,l2,l3,l4,l;//l1,l2,l3,l4,l分别表示四条边和一个对角线。 int i,t; scanf("%d",&t); while(t--) { sum1=sum2=0,x1=x2=x3=x4=y1=y2=y3=y4=inf;//初始坐标 for(i=0;i<4;i++) scanf("%lf %lf",&a[i],&b[i]),sum1+=a[i],sum2+=b[i]; sum1=sum1/4,sum2=sum2/4; for(i=0;i<4;i++)//更新坐标 { if(a[i]
        
         =sum2) x1=a[i],y1=b[i]; if(a[i]<=sum1&&b[i]
         
          =sum1&&b[i]>sum2) x3=a[i],y3=b[i]; if(a[i]>sum1&&b[i]<=sum2) x4=a[i],y4=b[i]; } if(x1==inf||x2==inf||x3==inf||x4==inf||y1==inf||y2==inf||y3==inf||y4==inf)//如果有一个坐标未更新 ，no { printf("No\n"); continue; } l1=pow(y1-y2,2)+pow(x1-x2,2);//和wa不同的地方 l2=sqrt(pow(y2-y4,2)+pow(x2-x4,2)); l3=sqrt(pow(y3-y4,2)+pow(x3-x4,2)); l4=pow(y1-y3,2)+pow(x1-x3,2);//同上 l=pow(y3-y2,2)+pow(x3-x2,2);//同上 if(sqrt(l1)==l2&&sqrt(l1)==l3&&sqrt(l1)==sqrt(l4)&&l1+l4==l) printf("Yes\n"); else printf("No\n"); } return 0; } 
         
        
       
      

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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
ac代码