HDU6055 Regular polygon（思路）_hdu 6055-优快云博客

本文介绍了一个算法问题，即如何计算平面上给定点集可以构成的不同正多边形的数量。通过输入一系列整数坐标点，程序将输出这些点能够组成的正多边形数目。文中提供了一段示例代码，该代码主要针对正方形的情况进行计算。

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

Sample Output

1
2

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);
    }
}