Hdu6055 Regular polygon（2017多校第2场）-优快云博客

本文介绍了一种通过枚举整数点集中的点来检测正多边形的算法，特别关注于正方形的检测。该算法适用于平面坐标系内的点集，并详细展示了如何通过检查对角线来确定是否存在构成正多边形的条件。

Regular polygon

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 986 Accepted Submission(s): 366

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

2017 Multi-University Training Contest - Team 2

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liuyiding

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题目的意思是给出n个整点，判有多少个正多边形

思路：只可能正方形，枚举对角线，判剩下的点存不存在

注意：算出来的点可能越界

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;
#define LL long long

const int INF = 0x3f3f3f3f;
const LL mod = 1e9 + 7;

struct point
{
    int x,y;
} p[505];

int n;
int vis[205][205];

int main()
{
    while(~scanf("%d",&n))
    {
        memset(vis,0,sizeof vis);
        for(int i=1; i<=n; i++)
        {
            scanf("%d%d",&p[i].x,&p[i].y);
            vis[p[i].x+100][p[i].y+100]=1;
        }
        int ans=0;
        for(int i=1; i<=n; i++)
        {
            for(int j=i+1; j<=n; j++)
            {
                point tmp;
                tmp.x=(p[i].x+p[j].x+p[j].y-p[i].y);
                tmp.y=(p[i].y+p[j].y+p[i].x-p[j].x);
                if(tmp.x%2!=0||tmp.y%2!=0) continue;
                tmp.x/=2,tmp.y/=2;
                if(tmp.x < -100 || tmp.x > 100 || tmp.y < -100 || tmp.y > 100 || !vis[tmp.x+100][tmp.y+100]) continue;
                tmp.x=(p[i].x+p[j].x+p[i].y-p[j].y);
                tmp.y=(p[i].y+p[j].y+p[j].x-p[i].x);
                if(tmp.x%2!=0||tmp.y%2!=0) continue;
                tmp.x/=2,tmp.y/=2;
                if(tmp.x < -100 || tmp.x > 100 || tmp.y < -100 || tmp.y > 100 || !vis[tmp.x+100][tmp.y+100]) continue;
                ans++;
            }
        }
        printf("%d\n",ans/2);
    }
    return 0;
}