HDU 6055 Regular polygon

Regular polygon

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1789 Accepted Submission(s): 702

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

题意：求能构成的正多边形的个数；但是由于输入的都是整数可知道构成的的一共都是正方形。
解题思路：先有两个点确定正方形的一条边，在有两点确定对角点，如果已知数据中同时存在两个对角点，即能形成正方形。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxx=510;
struct node
{
    int x,y;
    bool operator<(const node &text)const
    {
        if(x==text.x)
            return y<text.y;
        else
            return x<text.x;
    }
}nodes[maxx];
int main()
{
    int n;
    while(scanf("%d",&n)==1&&n)
    {
        int ans=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&nodes[i].x,&nodes[i].y);
        }
        sort(nodes,nodes+n);
        for(int i=0;i<n;i++)
        {
            for(int j=i+1;j<n;j++)
            {
                node tmp;
                tmp.x=nodes[i].x+nodes[i].y-nodes[j].y;
                tmp.y=nodes[i].y+nodes[j].x-nodes[i].x;
                if(!binary_search(nodes,nodes+n,tmp)) continue;
                tmp.x=nodes[j].x+nodes[i].y-nodes[j].y;
                tmp.y=nodes[j].y+nodes[j].x-nodes[i].x;
                if(!binary_search(nodes,nodes+n,tmp)) continue;
                ans++;
            }

        }
        printf("%d
",ans/2);
    }
    return 0;
}