解题报告——POJ 2002

Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 16908		Accepted: 6425

Description

A square is a 4-sided polygon whose sides have equal length and adjacent sides form 90-degree angles. It is also a polygon such that rotating about its centre by 90 degrees gives the same polygon. It is not the only polygon with the latter property, however, as a regular octagon also has this property. 

So we all know what a square looks like, but can we find all possible squares that can be formed from a set of stars in a night sky? To make the problem easier, we will assume that the night sky is a 2-dimensional plane, and each star is specified by its x and y coordinates. 

Input

The input consists of a number of test cases. Each test case starts with the integer n (1 <= n <= 1000) indicating the number of points to follow. Each of the next n lines specify the x and y coordinates (two integers) of each point. You may assume that the points are distinct and the magnitudes of the coordinates are less than 20000. The input is terminated when n = 0.

Output

For each test case, print on a line the number of squares one can form from the given stars.

Sample Input

4
1 0
0 1
1 1
0 0
9
0 0
1 0
2 0
0 2
1 2
2 2
0 1
1 1
2 1
4
-2 5
3 7
0 0
5 2
0

Sample Output

1
6
1
-------------------------------------------------------------------
这道题的原意应该是用哈希数组解决，不过我做了将近一天还是没有AC，不是WA就是RE，等技术娴熟了再来做吧~
这道题可以用二分法来做：将点集进行排序，然后用二分法进行查找。
大致思路：按照先x后y的顺序对点集进行排序，循环选择两个点，然后计算出第三、四个点可能存在的位置，用二分法查找这样的点是否存在，若存在则计数加一。最后根据同一正方形被重复计数的次数，将计数除以相应的次数。
代码如下：
------------------------------------------------------------------

#include <cstdio>
#include <algorithm>

using namespace std;

#define N 1001

struct Node
{
    int x, y;
}node[N];

Node temp;

bool cmp(Node X, Node Y)
{
    if(X.x == Y.x)
    {
        return X.y < Y.y;
    }
    return X.x < Y.x;
}

int main()
{
    int n, sum = 0;
    freopen("D:\test.txt", "r", stdin);
    while(scanf("%d", &n) != EOF && n)
    {
        sum = 0;
        for(int i = 0; i < n; i++)
        {
            scanf("%d%d", &node[i].x, &node[i].y);
        }
        sort(node, node + n, cmp);
        for(int i = 0; i < n; i++)
        {
            for(int j = i + 1; j < n; j++)
            {
                temp.x = node[j].x + node[i].y - node[j].y;
                temp.y = node[j].y + node[j].x - node[i].x;
                if(!binary_search(node, node + n, temp, cmp))
                {
                    continue;
                }
                temp.x = node[i].x + node[i].y - node[j].y;
                temp.y = node[i].y + node[j].x - node[i].x;
                if(binary_search(node, node + n, temp, cmp))
                {
                    sum++;
                }
            }
        }
        printf("%d
", sum / 2);
    }
}