HDU6055 Regular polygon（2017多校第二场）

最新推荐文章于 2018-08-12 20:59:07 发布

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本文介绍了一种通过暴力搜索算法来找出平面上由给定点构成的不同正多边形数量的方法。输入为若干组测试数据，每组包含一系列二维平面上的整数坐标点；输出则是基于这些点能构成的不同正多边形的数量。

Regular polygon

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1087 Accepted Submission(s): 405

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

论文？？？

被队友先入为主了，刚开始提了一下整数，后来想到死也没想出来，直接暴力了。

#include <iostream>
#include <cstring>

using namespace std;

int n,ans,a,b;

bool mapc[205][205];

struct point
{
    int x,y;
};

int judge(point a,point b)
{
    int x1 = a.x, y1 = a.y, x2 = b.x, y2 = b.y;
    int res = 0;
    int x3 = -y2+y1+x1 ,y3 = x2-x1+y1;
    int x4 = x3-x1+x2 ,y4 = y3-y1+y2;
    int x5 = 2*x1-x3 ,y5 = 2*y1-y3;
    int x6 = 2*x2-x4 ,y6 = 2*y2-y4;
    if( (x3<=100 && x3>=-100) && (y3<=100 && y3>=-100)
       && (x4<=100 && x4>=-100) && (y4<=100 && y4>=-100)
       && mapc[x3+100][y3+100] && mapc[x4+100][y4+100] )
        res++;
    if( (x5<=100 && x5>=-100) && (y5<=100 && y5>=-100)
       && (x6<=100 && x6>=-100) && (y6<=100 && y6>=-100)
       && mapc[x5+100][y5+100] && mapc[x6+100][y6+100] )
        res++;
    return res;
}

point p[505];
int main()
{
    while(cin>>n)
    {
        ans=0;
        memset(mapc,0,sizeof(mapc));
        for(int i=0;i<n;i++)
        {
            cin>>a>>b;
            p[i].x=a,p[i].y=b;
            mapc[a+100][b+100]=1;
        }
        for(int i=0;i<n;i++)
            for(int j=i+1;j<n;j++)
                ans+=judge(p[i],p[j]);
        cout<<ans/4<<endl;
    }

    return 0;
}