POJ 2002 Squares(几何 + 二分查找)

Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 11235		Accepted: 4096

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

 解题报告：这道题就是给出若干个点，判断最多能组成多少个正方形；先排序，然后枚举任意两点(x1,y1)(x2,y2)，则如果存在点(x1+y1-y2,y1-x1+x2)(x2+y1-y2,y2-x1+x2)则它们能构成一个正方形(这里方向是确定的，否则还有一种可能)。再利用二分查询是否存在这两个点，存在则ans ++;最后的ans要/2，因为正方形都重复算了一次。有的人用的hash，以后再利用这种方法做

代码如下：自己编了一个，看见网上都用STL写，我也稍作改变写了一个

1>自己编的二分查找2188MS

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAX = 1010;
struct Point
{
    int x;
    int y;
}p[MAX];
bool cmp(const Point &a, const Point &b)//把x从小到大排序，若x相等则再按y从小到大排序
{
    if (a.x == b.x) return a.y < b.y;
    else return a.x < b.x;
}
bool Binsearch(int low, int high, Point a)//二分查找算法
{
    int mid;
    if (low <= high)
    {
        mid =low + (high - low) / 2;
        if (a.x == p[mid].x && a.y == p[mid].y)//找到
        {
            return true;
        }
        if (a.x > p[mid].x ||(a.x == p[mid].x && a.y > p[mid].y))//判断的时候先判断a.x与p[mid].x;再判断y的大小
        {
            return Binsearch(mid + 1, high, a);//进入右半部分
        }
        else
        {
            return Binsearch(low, mid - 1, a);//进入左半部分
        }
    }
    return false;
}
int main()
{
    int n, i, ans, j;
    while (scanf("%d", &n) != EOF && n)
    {
        for (i = 0; i < n; ++i)
        {
            scanf("%d%d", &p[i].x, &p[i].y);
        }
        sort(p, p + n, cmp);
        ans = 0;
        for (i =0; i < n; ++i)
        {
            for (j = i + 1; j < n; ++j)
            {
                Point temp;
                temp.x = p[i].x + p[i].y - p[j].y;
                temp.y = p[i].y - p[i].x + p[j].x;
                if (!Binsearch(0, n - 1, temp))
                {
                    continue;
                }
                temp.x = p[j].x + p[i].y - p[j].y;
                temp.y = p[j].y - p[i].x + p[j].x;
                if (!Binsearch(0, n - 1, temp))
                {
                    continue;
                }
                ans ++;//两点同时满足才存在
            }
        }
        printf("%d\n", ans / 2);
    }
    return 0;
}

2>用STL写的，第一次用STL，代码短而且时间也快1625MS

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAX = 1010;
struct Point
{
    int x;
    int y;
}p[MAX];
bool cmp(const Point &a, const Point &b)//把x从小到大排序，若x相等则再按y从小到大排序
{
    if (a.x == b.x) return a.y < b.y;
    else return a.x < b.x;
}
int main()
{
    int n, i, ans, j;
    while (scanf("%d", &n) != EOF && n)
    {
        for (i = 0; i < n; ++i)
        {
            scanf("%d%d", &p[i].x, &p[i].y);
        }
        sort(p, p + n, cmp);
        ans = 0;
        for (i =0; i < n; ++i)
        {
            for (j = i + 1; j < n; ++j)
            {
                Point temp;
                temp.x = p[i].x + p[i].y - p[j].y;
                temp.y = p[i].y - p[i].x + p[j].x;
                if(!binary_search(p, p + n, temp, cmp))
                {
                    continue;
                }
                temp.x = p[j].x + p[i].y - p[j].y;
                temp.y = p[j].y - p[i].x + p[j].x;
                if(!binary_search(p, p + n, temp, cmp))
                {
                    continue;
                }
                ans ++;
            }
        }
        printf("%d\n", ans / 2);
    }
    return 0;