Regular polygon 数学几何

最新推荐文章于 2020-05-14 15:07:49 发布

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最新推荐文章于 2020-05-14 15:07:49 发布

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分类专栏：数学（几何）

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本文介绍了一道编程题目，该题目要求在给定的二维平面上找到由整数点构成的不同正多边形的数量。通过枚举任意两点并检查是否存在另外两点以形成正方形的方式解决此问题。提供了完整的AC代码实现。

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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

4
0 0
0 1
1 0
1 1
6
0 0
0 1
1 0
1 1
2 0
2 1

Sample Output

1
2

Hint

题意

题解：

只能构成正方形暴力枚举任意两个点判断即可

AC代码

#include <cstdio>
#include <iostream>
#include <queue>
#include <map>
#include <vector>
#include <cmath>
#include <set>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<int ,int> P;
const int N = 500;
int vis[N][N];
struct node{
    int x,y;
}p[N];
int main(){
    int n;
    while (scanf("%d",&n)!=EOF){
            memset(vis,0,sizeof(vis));
        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] = 1;
        }
        int ans = 0;
        for (int i = 0; i < n; ++i){
            for (int j = i+1; j < n; ++j){
                 node a = p[i];node b = p[j];
                 int dx = a.x-b.x;
                 int dy = a.y-b.y;
                 if (a.x+dy>=0&&a.y-dx>=0&&b.y-dx>=0&&b.x+dy>=0&&vis[a.x+dy][a.y-dx]&&vis[b.x+dy][b.y-dx]){
                    ans++;
                 }
                 if (a.x-dy>=0&&a.y+dx>=0&&b.y+dx>=0&&b.x-dy>=0&&vis[a.x-dy][a.y+dx]&&vis[b.x-dy][b.y+dx]){
                    ans++;
                 }
            }
        }
        printf("%d\n",ans/4);

    }
    return 0;
}