POJ 2002 Squares

二分。。。。

Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 14530		Accepted: 5488

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

Source

Rocky Mountain 2004 

枚举两个点计算另外俩个，二分查找。。。。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

struct node
{
    int x,y;
}p[1100],p1,p2;

int n,cnt;

bool cmp(node a,node b)
{
    if(a.x!=b.x) return a.x<b.x;
    return a.y<b.y;
}

bool findp(node a)
{
    int s=0,t=n-1,m;
    int X=a.x,Y=a.y;
    while(s<=t)
    {
        m=(s+t)/2;
        if(p.x==X&&p.y==Y) return true;
        if(p.x<X||(p.x==X&&p.y<Y)) s=m+1;
        else t=m-1;
    }
    return false;
}

int main()
{
    while(scanf("%d",&n)!=EOF&&n)
    {
        cnt=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&p.x,&p.y);
        }
        sort(p,p+n,cmp);
        for(int i=0;i<n;i++)
        {
            for(int j=i+1;j<n;j++)
            {
                p1.x=p.x+p[j].y-p.y; p1.y=p.y+p.x-p[j].x;
                p2.x=p[j].x+p[j].y-p.y; p2.y=p[j].y+p.x-p[j].x;
                if(!findp(p1)) continue;
                if(findp(p2)) cnt++;
            }
        }
        printf("%d ",cnt/2);
    }
    return 0;
}

* This source code was highlighted by YcdoiT. ( style: Codeblocks )