HDU-6055-Regular polygon_hdu6055 java-优快云博客

本文链接：https://blog.youkuaiyun.com/chenyang890/article/details/77248996

本文介绍了一个算法问题：给定二维平面上的整数坐标点集，任务是找出这些点能构成的不同正多边形的数量。输入包括多个测试用例，每个用例给出一系列点坐标；输出则是每个用例中可以构成的不同正多边形数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Regular polygon

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2546 Accepted Submission(s): 1018

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

解题代码

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
struct node
{
    int x,y;
}ch[505];
int vis[205][205];
int main()
{
    int N;
    while(scanf("%d",&N)!=EOF)
    {
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=N;i++)
        {
            scanf("%d%d",&ch[i].x,&ch[i].y);
            ch[i].x+=100;
            ch[i].y+=100;
            vis[ch[i].x][ch[i].y]=1;
        }
        int count1=0;
        for(int i=1;i<=N;i++)
            for(int j=i+1;j<=N;j++)
        {
            int x1=ch[i].x;
            int y1=ch[i].y;
            int x2=ch[j].x;
            int y2=ch[j].y;
            int dx=ch[i].x-ch[j].x;
            int dy=ch[i].y-ch[j].y;
            if(x1+dy>=0&&x1+dy<=200&&y1-dx>=0&&y1-dx<=200&&x2+dy>=0&&x2+dy<=200&&y2-dx>=0&&y2-dx<=200)
            {
                if(vis[x1+dy][y1-dx]==1&&vis[x2+dy][y2-dx]==1)
                    count1++;
            }
            if(x1-dy>=0&&x1-dy<=200&&y1+dx>=0&&y1+dx<=200&&x2-dy>=0&&x2-dy<=200&&y2+dx>=0&&y2+dx<=200)
            {
                if(vis[x1-dy][y1+dx]==1&&vis[x2-dy][y2+dx]==1)
                    count1++;
            }
        }
        printf("%d\n",count1/4);
    }
    return 0;
}