HDU6055 Regular polygon（思路）

Regular polygon

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 111    Accepted Submission(s): 40

Problem Description

On a two-dimensional plane, give you n integer points. Your task is to figure out how many different regular polygon these points can make.

 

Input

The input file consists of several test cases. Each case the first line is a numbers N (N <= 500). The next N lines ,each line contain two number Xi and Yi(-100 <= xi,yi <= 100), means the points’ position.(the data assures no two points share the same position.)

 

Output

For each case, output a number means how many different regular polygon these points can make.

 

Sample Input

40 00 11 01 160 00 11 01 12 02 1

 

Sample Output

12

 

Source

由于给出的点都是整数而且要组合成正多边形，只可能组合出正方形，接下来枚举正方形就行了

#include<bits/stdc++.h>#define mem(a,b) memset(a,b,sizeof(a))using namespace std;const int N=505;struct node{    int x,y;}p[N];bool vis[N][N];int solve(node a,node b){    int ans=0;    int x=a.x-b.x;    int y=a.y-b.y;    if(a.x+y>=0&&a.y-x>=0&&b.x+y>=0&&b.y-x>=0&&vis[a.x+y][a.y-x]&&vis[b.x+y][b.y-x])    ans++;    if(a.x-y>=0&&a.y+x>=0&&b.x-y>=0&&b.y+x>=0&&vis[a.x-y][a.y+x]&&vis[b.x-y][b.y+x])    ans++;    return ans;}int main(){    int n;    while(~scanf("%d",&n))    {        mem(vis,false);        for(int i=0;i<n;i++)        {            scanf("%d%d",&p[i].x,&p[i].y);            p[i].x+=200;            p[i].y+=200;            vis[p[i].x][p[i].y]=true;        }        int ans=0;        for(int i=0;i<n;i++)        {            for(int j=i+1;j<n;j++)            {                ans+=solve(p[i],p[j]);            }        }       printf("%d\n",ans/4);    }}