hdu5365 Run

Link：http://acm.hdu.edu.cn/showproblem.php?pid=5365

Run

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 529    Accepted Submission(s): 231

Problem Description

AFA is a girl who like runing.Today,he download an app about runing .The app can record the trace of her runing.AFA will start runing in the park.There are many chairs in the park,and AFA will start his runing in a chair and end in this chair.Between two chairs,she running in a line.she want the the trace can be a regular triangle or a square or a regular pentagon or a regular hexagon.
Please tell her how many ways can her find.
Two ways are same if the set of chair that they contains are same.

 

Input

There are multiply case.
In each case,there is a integer n(1 < = n < = 20)in a line.
In next n lines,there are two integers xi,yi(0 < = xi,yi < 9) in each line.

 

Output

Output the number of ways.

 

Sample Input

  

   4
0 0
0 1
1 0
1 1
  

 

Sample Output

  

   1
  

 

Source

BestCoder Round #50 (div.2)

 

编程思想：由于输入数据x、y坐标是整数，对应的点连起来不可能是正三角形、正五边形和正六边形，只可能是正四边形。根据正四边形的判定定理：四条边相等且有一个角是直角的四边形是正四边形。通过枚举每四个点所构成的六条边，若这四个点能构成正四边形，则其所构成的六条边中有四条是正四边形的边，另外两条是正四边形的对角线，且对角线比边长，故通过从小到大排序即可判断前4条边为正四边形的边，后两条为对角线。根据正四边形的边相等和正四边形的对角线与正四边形的边构成直角三角形的勾股关系，即可判定是否为正四边形。

AC code：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define LL long long 
#define exp 1e-10
#define MAXN 1000010
using namespace std;
const int INF=0x3f3f3f3f;
const int N = 100005;
const int mod = 1000000007;
int n,i,j,k,l,ans;
int x[33],y[33]; 
int e[33];
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
    	for(i=1;i<=n;i++)
    	{
    		scanf("%d%d",&x[i],&y[i]);
		}
		ans=0;
		for(i=1;i<=n;i++)
		{
			for(j=i+1;j<=n;j++)
			{
				for(k=j+1;k<=n;k++)
				{
					for(l=k+1;l<=n;l++)
					{
						e[1]=(x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]);
						e[2]=(x[i]-x[k])*(x[i]-x[k])+(y[i]-y[k])*(y[i]-y[k]);
						e[3]=(x[i]-x[l])*(x[i]-x[l])+(y[i]-y[l])*(y[i]-y[l]);
						e[4]=(x[j]-x[k])*(x[j]-x[k])+(y[j]-y[k])*(y[j]-y[k]);
						e[5]=(x[j]-x[l])*(x[j]-x[l])+(y[j]-y[l])*(y[j]-y[l]);
						e[6]=(x[k]-x[l])*(x[k]-x[l])+(y[k]-y[l])*(y[k]-y[l]);
						sort(e+1,e+7);
						if(e[1]==e[2]&&e[2]==e[3]&&e[3]==e[4]&&e[3]+e[4]==e[5])
						{
							ans++;
						}
					}
				}
			}
		}
		printf("%d\n",ans);
	}
}