POJ 2002 Squares

最新推荐文章于 2020-04-17 15:27:14 发布

转载最新推荐文章于 2020-04-17 15:27:14 发布 · 248 阅读

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原文链接：http://www.cnblogs.com/CKboss/p/3350865.html

本文介绍了一种通过枚举和二分查找的方法来解决寻找由星星构成的正方形的问题。输入为一组星星坐标，输出为可能形成的正方形数量。源代码使用C++实现，并详细解释了算法流程。

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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\n",cnt/2);
    }
    return 0;
}