poj 2002 Squares

题目链接：http://poj.org/problem?id=2002

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.

题意：二维平面给出一些点，计算以这些点为顶点的正方形的个数。

解法：哈希。因为n为1000，坐标大小高达20000，普通查找肯定不行的，我们得用哈希压缩坐标。

　　   然后枚举正方形的一条边上的两个顶点，可以计算出剩余的两个顶点坐标，然后哈希表查找是否有这样的正方形，统计个数，然后除以4即可（因为枚举了每个正方形的每条边）。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<cmath>
 6 #include<algorithm>
 7 #define inf 0x7fffffff
 8 using namespace std;
 9 typedef long long ll;
10 const int maxn=1000+10;
11 const int mod=100007;
12 struct node
13 {
14     int x,y;
15     int next;
16 }edge[maxn];
17 int head[mod],edgenum;
18 int xx[maxn],yy[maxn];
19 void insert_hash(int x,int y)
20 {
21     int k=(x*x+y*y)%mod;
22     edge[edgenum].x=x ;
23     edge[edgenum].y=y ;
24     edge[edgenum].next=head[k] ;
25     head[k]=edgenum++ ;
26 }
27 int search_hash(int x,int y)
28 {
29     int k=(x*x+y*y)%mod;
30     for (int i=head[k] ;i!=-1 ;i=edge[i].next)
31     {
32         if (x==edge[i].x && y==edge[i].y) return true;
33     }
34     return false;
35 }
36 int main()
37 {
38     int n;
39     while (scanf("%d",&n)!=EOF && n)
40     {
41         memset(head,-1,sizeof(head));
42         edgenum=0;
43         for (int i=0 ;i<n ;i++)
44         {
45             scanf("%d%d",&xx[i],&yy[i]);
46             insert_hash(xx[i],yy[i]);
47         }
48         int count=0;
49         for (int i=0 ;i<n ;i++)
50         {
51             for (int j=i+1 ;j<n ;j++)
52             {
53                 int x=xx[i]+(yy[j]-yy[i]);
54                 int y=yy[i]-(xx[j]-xx[i]);
55                 int x2=xx[j]-(yy[i]-yy[j]);
56                 int y2=yy[j]+(xx[i]-xx[j]);
57                 if (search_hash(x,y) && search_hash(x2,y2)) count ++ ;
58             }
59         }
60         for (int i=0 ;i<n ;i++)
61         {
62             for (int j=i+1 ;j<n ;j++)
63             {
64                 int x=xx[i]-(yy[j]-yy[i]);
65                 int y=yy[i]+(xx[j]-xx[i]);
66                 int x2=xx[j]+(yy[i]-yy[j]);
67                 int y2=yy[j]-(xx[i]-xx[j]);
68                 if (search_hash(x,y) && search_hash(x2,y2)) count ++ ;
69             }
70         }
71         printf("%d\n",count/4);
72     }
73     return 0;
74 }
View Code

 

 

后续：感谢大牛提出宝贵的意见。。。

posted @ 2014-12-05 20:58  huangxf  阅读(269)  评论(0)  编辑  收藏  举报