poj 2002 (hash map)

Squares

Time Limit: 3500 MSMemory Limit: 65536 KB

64-bit integer IO format: %I64d , %I64u Java class name: Main

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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

题意是给你n个点,问你能找到多少个正方形.

思路:对任意两个点遍历,判断其能组成正方形的点的坐标是否存在.这里就需要在输入点的时候对点hash处理,我这里用的是map,但后来发现超时了,将map的调用改为用count()后就刚好能过.

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<map>
using namespace std;
const int MAX = 20009;

struct Point{
    int x,y;
}point[1009];

long long solve(int x,int y)
{
    return x+MAX+(y+MAX)*(MAX*2);
}

int main()
{
    int pointnum;
    while(scanf("%d",&pointnum) != EOF){
        if(pointnum == 0)
        return 0;
        map<long long,int>mapp;
        long long ans = 0;
        int x1,x2,y1,y2;
        for(int i = 0;i < pointnum;i++){
            scanf("%d%d",&point[i].x,&point[i].y);
            long long number = solve(point[i].x,point[i].y);
            mapp[number] = 2;
        }
        for(int i = 0;i < pointnum;i++){
            for(int j = i+1;j < pointnum;j++){
                x1 = point[i].x;
                y1 = point[i].y;
                x2 = point[j].x;
                y2 = point[j].y;
                long long a = x1-(y2-y1)+MAX+(2*MAX)*(MAX+y1+x2-x1);
                long long b = x2+(y1-y2)+MAX+(2*MAX)*(MAX+y2-(x1-x2));
                if(mapp.count(a) && mapp.count(b)){
                    ans++;
                }
                a = x2-(y1-y2)+MAX+(2*MAX)*(MAX+y2+(x1-x2));
                b = x1+y2-y1+MAX+(2*MAX)*(MAX+y1-(x2-x1));
                if(mapp.count(a) && mapp.count(b)){
                    ans++;
                }
            }
        }

        printf("%lld\n",ans/4);
        mapp.clear();
    }
    return 0;
}