HDU 6055 Regular polygon

最新推荐文章于 2018-07-16 16:23:36 发布

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本文介绍了一个算法问题：如何确定给定点集能在二维平面上构成多少种不同的正多边形。输入为一系列整数坐标点，输出则是能构成的不同正多边形的数量。通过比较和搜索算法来解决这一几何问题。

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Regular polygon
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 523 Accepted Submission(s): 201

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

#include<cstdio>
#include<algorithm>
#include <vector>
using namespace std;
int n;
int sum;
struct note
{
    int x,y;
} a[505];
note tt;
int cmp(note a,note b)
{
    if(a.x==b.x)
        return a.y<b.y;
    return a.x<b.x;
}

int main()
{

    while(~scanf("%d",&n))
    {
        sum=0;
        for(int i=0; i<n; i++)
        {
            scanf("%d %d",&a[i].x,&a[i].y);
        }
        sort(a,a+n,cmp);
        for(int i=0; i<n; i++)
        {
            for(int j=i+1; j<n; j++)
            {
                tt.x=a[i].x-a[j].y+a[i].y;
                tt.y=a[i].y+a[j].x-a[i].x;
                if(!binary_search(a,a+n,tt,cmp)) continue;

                tt.x=a[j].x-a[j].y+a[i].y;
                tt.y=a[j].y+a[j].x-a[i].x;
                if(!binary_search(a,a+n,tt,cmp)) continue;

                sum++;
            }
        }
        printf("%d\n",sum/2);
    }
    return 0;
}