【STL】Squares

Squares

http://poj.org/problem?id=2002

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

41 00 11 10 090 01 02 00 21 22 20 11 12 14-2 53 70 05 20

Sample Output

161

题意就是找出能组成正方形的点的个数。

学习了下STL中的binary_search

 #include <algorithm> bool binary_search( forward_iterator start, forward_iterator end, const TYPE& val ); bool binary_search( forward_iterator start, forward_iterator end, const TYPE& val, Comp f );

如果找到了val, binary_search()返回true, 否则返回false. 如果函数f被指定, 它将代替比较元素值的函数operator < .

#include<cstdio>#include<algorithm>using namespace std;struct point{    int x,y;}node[1005];bool cmp(const struct point &a,const struct point &b){    return a.x==b.x?(a.y<b.y):(a.x<b.x);}int main(){    int n;    struct point temp;    for(;(~scanf("%d",&n))&&n;)    {        int summ=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[i].x+node[i].y-node[j].y;                temp.y=node[i].y-node[i].x+node[j].x;                if(!binary_search(node,node+n,temp,cmp))                   continue;                temp.x=node[j].x+node[i].y-node[j].y;                temp.y=node[j].y-node[i].x+node[j].x;                if(!binary_search(node,node+n,temp,cmp))                   continue;                summ++;            }        }        printf("%d\n",summ/2);    }    return 0;}