HDU6055 Regular polygon(计算几何）

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

Sample Input

Sample Output

题意：

　　给定n个整数点，问能组成多少正n边形。

思路：

　　因为给出的是整数点，所以只可能是正方形

　　可以枚举正方形的对角线，因为有两条对角线，最后答案要/2

　　也可以枚举正方形的边，因为有四条边，答案要/4

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

#define MAX 2000

using namespace std;

struct node
{
    int x,y;
} p[MAX];

int vis[MAX][MAX];

int solve(int node1,int node2)
{
    int x = p[node1].x - p[node2].x;
    int y = p[node1].y - p[node2].y;
    int ans=;
    if(p[node1].x-y>= && p[node1].y+x>= && p[node2].x-y>= && p[node2].y+x>= )//判断成正方形的条件
        if(vis[p[node1].x-y][p[node1].y+x] && vis[p[node2].x-y][p[node2].y+x])//判断两个点是否存在
            ans++;
    if(p[node1].x+y>= && p[node1].y-x>= && p[node2].x+y>= && p[node2].y-x>= )
        if( vis[p[node1].x+y][p[node1].y-x] && vis[p[node2].x+y][p[node2].y-x])
            ans++;
    return ans;
}

int main()
{
    int n,x,y;
    while(scanf("%d",&n)!=EOF)
    {
        int ans = ;
        memset(vis,,sizeof(vis));
        //vis数组用来查看是否存在，结构体p用来存所有的点
        for(int i=; i<n; i++)
        {
            scanf("%d%d",&x,&y);
            vis[x+][y+] = ;//注意+200是为了将负数化为正数
            p[i].x = x+;
            p[i].y = y+;
        }
        //枚举所有的点
        for(int i=; i<n; i++)
        {
            for(int j=i+; j<n; j++)
            {
                ans += solve(i,j);
            }
        }

        ans /= ;//因为枚举的是边长，所以最后除以4

        printf("%d\n",ans);
    }
    return ;
}