hdu 5365 计算几何 给几个点判断是否为正方形

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Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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

40 00 11 01 1

 

Sample Output

1

最后题目意思就是给你一系列点，问能组成正三角形，正四边形，正五边形，正六边形的个数（如果使用的点相同，则视为一种），实际上由于正三角形正五边形，正六边形不可能都由整数点（即横坐标和纵坐标都是整数）构成，故我们只用判断能组成多少个正方形即可。

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<queue>#include<stack>#include<math.h>#include<vector>#include<map>#include<set>#include<stdlib.h>#include<cmath>#include<string>#include<algorithm>#include<iostream>using namespace std;struct point{    int  x,y;} pp[110];int s[6];bool judge(int a,int b,int c,int d){    int cnt=0;    s[cnt++]=(pp[b].x-pp[a].x)*(pp[b].x-pp[a].x)+(pp[b].y-pp[a].y)*(pp[b].y-pp[a].y);    s[cnt++]=(pp[c].x-pp[a].x)*(pp[c].x-pp[a].x)+(pp[c].y-pp[a].y)*(pp[c].y-pp[a].y);    s[cnt++]=(pp[d].x-pp[a].x)*(pp[d].x-pp[a].x)+(pp[d].y-pp[a].y)*(pp[d].y-pp[a].y);    s[cnt++]=(pp[c].x-pp[b].x)*(pp[c].x-pp[b].x)+(pp[c].y-pp[b].y)*(pp[c].y-pp[b].y);    s[cnt++]=(pp[d].x-pp[b].x)*(pp[d].x-pp[b].x)+(pp[d].y-pp[b].y)*(pp[d].y-pp[b].y);    s[cnt++]=(pp[d].x-pp[c].x)*(pp[d].x-pp[c].x)+(pp[d].y-pp[c].y)*(pp[d].y-pp[c].y);    sort(s,s+6);//以上代码就是在求这四个点构成的边的长度，如果是正方形，那么排序后前四个大小一样，最后后两个大小一样    if(s[0]==s[1]&&s[2]==s[1]&&s[2]==s[3]&&s[4]==s[5]&&s[4]!=s[3])return true;    return false;}int main(){    int  t,i,j,n,k,l,cnt;    while(scanf("%d",&n)!=EOF)    {        cnt=0;        for(i=0; i<n; i++)            scanf("%d%d",&pp[i].x,&pp[i].y);        for(i=0; i<n; i++)            for(j=i+1; j<n; j++)                for(k=j+1; k<n; k++)                    for(l=k+1; l<n; l++)                        if(judge(i,j,k,l))cnt++;       printf("%d\n",cnt);    }    return 0;}