poj 2002 Squares

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poj 2002 Squares
Squares
Time Limit: 3500MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u

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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

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

Sample Output

161

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题意:给你若干个点,让你找出这些点可以组成正方形的个数。

分析:如果暴力的话时间复杂度为O(n^4)肯定会 TLE  ,这里使用了hash查找。不过也有二分做的。。

Source Code

Problem: User: sdau_09_zysMemory: 1352 KB Time: 969 MSLanguage: G++ Result: AcceptedPublic: 

#include <iostream>#include <iomanip>#include <cstdio>#include <cstring>using namespace std;struct Point{    int x,y;}p[1001];const int size=22000;/**  * hash表为啥要开到22000?  * 题目中给的两点的距离小于20000,  * 到原点的距离完全可能大于20000,  * 这里有猜测的意味。***/struct Node{    Node():next(0){}//对next初始化    int x,y;    Node * next;}hash[size];//拉链法的hash表bool findPoint(int x,int y){    int key=(x*x+y*y)%size;    Node *pt=&hash[key];    while(pt->next)    {        if(pt->next->x==x&&pt->next->y==y)return true;        pt=pt->next;    }    return false;}int main(){    int n,i,j,key,x,y,x1,y1,x2,y2,ans;    Node* tem;    Node* pt;    while(scanf("%d",&n),n)    {        ans=0;        memset(hash,0,sizeof hash);        for(i=0;i<n;i++)        {            scanf("%d %d",&p[i].x,&p[i].y);            tem=new Node;            tem->x=p[i].x;tem->y=p[i].y;            key=(p[i].x*p[i].x+p[i].y*p[i].y)%size;            //用到原点的距离作为hash函数            pt=&hash[key];            while(pt->next)            {                pt=pt->next;            }            pt->next=tem;        }        for(i=0;i<n-1;i++)        {            for(j=i+1;j<n;j++)            {                x=p[i].x-p[j].x;                y=p[i].y-p[j].y;                //斜下方的正方形                x1=p[i].x+y;                y1=p[i].y-x;                x2=p[j].x+y;                y2=p[j].y-x;                if(findPoint(x1,y1)&&findPoint(x2,y2))ans++;                //斜上方的正方形                x1=p[i].x-y;                y1=p[i].y+x;                x2=p[j].x-y;                y2=p[j].y+x;                if(findPoint(x1,y1)&&findPoint(x2,y2))ans++;            }        }        printf("%d\n",ans>>2);//为啥要除以4呢?是因为正方形的4个边都会被枚举一次    }}

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