HDU 6055 Regular polygon

Regular polygon

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1529    Accepted Submission(s): 597

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

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

 

 

Source

2017 Multi-University Training Contest - Team 2

 

 

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/*
* @Author: Lyucheng
* @Date:   2017-07-27 14:26:58
* @Last Modified by:   Lyucheng
* @Last Modified time: 2017-07-28 15:43:15
*/
/*
 题意：给你一个点阵，让你找多边形的个数，因为点都是整数所以只可能是正方形
*/


#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>

#define MAXN 5005
#define MAXM 2005
#define EXP 1e-6
using namespace std;

struct Point{
    int x,y;
    Point(){}
    Point(int _x,int _y){
        x=_x;
        y=_y;
    }
    bool operator < (const Point & other) const{
        if(x==other.x)
            return y<other.y;
        return x<other.x;
    }
}point[MAXN];
int n;
bool vis[MAXM][MAXM];

bool ok(Point a,Point b){//对点
    double ax=(a.x+a.y+b.x-b.y)*1.0/2;
    double ay=(-a.x+a.y+b.x+b.y)*1.0/2;
    double bx=(a.x-a.y+b.x+b.y)*1.0/2;
    double by=(a.x+a.y-b.x+b.y)*1.0/2;

    if(ax-(int)ax>EXP||bx-(int)bx>EXP||ay-(int)ay>EXP||by-(int)by>EXP)
        return false;
    if(ax<0||ay<0||bx<0||by<0)
        return false;
    
    if(vis[(int)ax][(int)ay]==true&&vis[(int)bx][(int)by]==true)
        return true;
    return false;
}

inline void init(){
    memset(vis,false,sizeof vis);
}

int main(){
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    while(scanf("%d",&n)!=EOF){
        init();
        for(int i=0;i<n;i++){
            scanf("%d%d",&point[i].x,&point[i].y);
            point[i].x+=100;
            point[i].y+=100;
            vis[point[i].x][point[i].y]=true;
        }
        sort(point,point+n);
        int res=0;
        for(int i=0;i<n;i++){
            for(int j=i+1;j<n;j++){
                if(ok(point[i],point[j])==true){
                    res++;
                }
            }
        }
        printf("%d\n",res/2);
    }
    return 0;
}