poj 2002 Squares

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Squares

Time Limit: 3500MS Memory Limit: 65536KTotal Submissions: 15879 Accepted: 6009

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

题目大意：

给你在纸面的n个点的x坐标以及y坐标，在这些点中能找到几个正方形。

这几天在做hash，见到这个题目还是没有太多的思路，后来看别人博客发现是hash每个点到（0，0）的距离而进行统计的。

也就是说把每个点的坐标都先统计进一个结构（先后用的链表以及vector的写法）中，按照到原点的距离来分类。

然后统计的时候是在1~n以及i+1~n中统计每两个点符合条件的另外两个点，几何推一下很容易可以得出来

如何避免冲突，下面有两种写法：

方法一：（链表）

也算复习一下学过的数据结构

个人感觉太过冗长....... 代码水平太差....... 感觉一不小心就会写搓

#include<iostream>#include<stdio.h>#include<cstring >using namespace std;#define mod 1999   //最好定义成小于2*n的最大素数struct node{    int x;    int y;};struct  HashTable{    int x;    int y;    HashTable *next;    HashTable()    {        next=0;    }};node pos[1001];HashTable *hash[mod];void insert(int k){    int key=(pos[k].x*pos[k].x+pos[k].y*pos[k].y)%mod;    if(!hash[key])    {        HashTable* temp=new HashTable;        temp->x=pos[k].x;        temp->y=pos[k].y;        hash[key]=temp;    }    else    {        HashTable* temp=hash[key];  //这一步之前错了，之前申请的不是一个链表，是一个hashtable结点，值存了一个元素        while(temp->next)            temp=temp->next;        temp->next=new HashTable;        temp->next->x=pos[k].x;        temp->next->y=pos[k].y;    }    return;}bool find(int a,int b){    int key=(a*a+b*b)%mod;    if(!hash[key]) return false;    else    {        HashTable *temp=hash[key];        while(temp)        {            if(temp->x==a && temp->y==b) return true;            temp=temp->next;        }    }    return false;}int main (){    int n;    while(cin>>n)    {        if(n==0) break;        memset(hash,0,sizeof(hash));        for(int i=1; i<=n; i++)        {            scanf("%d%d",&pos[i].x,&pos[i].y);            insert(i);        }        int num=0;        for(int i=1;i<=n;i++)         for(int j=i+1;j<=n;j++)         {             int a=pos[j].x-pos[i].x;             int b=pos[j].y-pos[i].y;             int x3=pos[i].x-b;             int y3=pos[i].y+a;             int x4=pos[j].x-b;             int y4=pos[j].y+a;             if(find(x3,y3) && find(x4,y4))             printf("* %d %d * %d %d*\n",x3,y3,x4,y4),             num++;             x3=pos[i].x+b;             y3=pos[i].y-a;             x4=pos[j].x+b;             y4=pos[j].y-a;             if(find(x3,y3) && find(x4,y4))             printf("* %d %d * %d %d*\n",x3,y3,x4,y4),             num++;         }//         for(int i=0;i<=100;i++)//         {//              if(hash[i])//              {//                   HashTable *temp=hash[i];//                   cout<<i<<" **";//                   while(temp)//                   {//                         cout<<temp->x<<" "<<temp->y<<"**";//                         temp=temp->next;//                   }//                   cout<<endl;//              }////         }         cout<<num/4<<endl;  //同一个正方形枚举了4次    }}/*160 10 20 30 41 11 21 31 42 12 22 32 43 13 23 33 4*/

方法二：vector写法

简单多了，明了多了，好用........

#include<iostream>#include<stdio.h>#include<vector>using namespace std;#define mod 1999struct point{    int x,y;};point p[1005];vector<int> hash[2000];bool find(int a,int b){    int key=(a*a+b*b)%mod;    for(vector<int>::size_type j = 0; j < hash[key].size(); j++ )    {        if(p[ hash[key][j] ].x==a && p[ hash[key][j] ].y==b) return true;    }    return false;}void insert(int k){    int key=( p[k].x*p[k].x + p[k].y*p[k].y )%mod;    hash[key].push_back(k);}int main (){    int n;    while(~scanf("%d",&n))    {        if(n==0) break;        int num=0;        for(int i=0;i<1999;i++)        hash[i].clear();        for(int i=1;i<=n;i++)        {            scanf("%d%d",&p[i].x,&p[i].y);            insert(i);        }       for(int i=1;i<=n;i++)         for(int j=i+1;j<=n;j++)         {             int a=p[j].x-p[i].x;             int b=p[j].y-p[i].y;             int x3=p[i].x-b;             int y3=p[i].y+a;             int x4=p[j].x-b;             int y4=p[j].y+a;             if(find(x3,y3) && find(x4,y4))             //printf("* %d %d * %d %d *\n",x3,y3,x4,y4),             num++;             x3=p[i].x+b;             y3=p[i].y-a;             x4=p[j].x+b;             y4=p[j].y-a;             if(find(x3,y3) && find(x4,y4))             //printf("* %d %d * %d %d *\n",x3,y3,x4,y4),             num++;         }         cout<<num/4<<endl;    }}

wa了好几发是因为......每个case里面没有清空vector........多case要注意清空和初始化

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