Squares（枚举+set 查找）

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

题意：给出n组坐标，判断这些坐标能组成的正方形的个数。

思路：参考某大神的想法，先枚举两个点，然后利用公式表示出另外两个点，判断这两个点是否在这n组坐标中,其中查找另两个坐标用的set容器。

已知 (x1,y1)(x2,y2);

则：x3 = x1+(y1-y2); y3 = y1 -(x1-x2);

     x4 = x2 +(y1-y2);y4 = y2 -(x1-x2);

或：x3 = x1 -(y1-y2);y3 = y1+(x1-x2);

     x4 = x2 -(y1-y2);y4 = y2 +(x1-x2);

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <set>
 5 const int N=100000;
 6 const int maxn=1010;
 7 
 8 using namespace std;
 9 struct node
10 {
11     long long  x;
12     long long  y;
13 } p[maxn];
14 int main()
15 {
16     int n;
17     while(~scanf("%d",&n)&&n)
18     {
19         int ans = 0;
20         set<long long>tt;
21         for (int i = 0; i < n; i ++)
22         {
23             scanf("%lld %lld",&p[i].x,&p[i].y);
24             tt.insert(p[i].x*N+p[i].y);//将坐标转化成一个数，坐标不同，则这个数就不同
25         }
26         for (int i = 0; i < n; i ++)
27         {
28             int x1 = p[i].x;
29             int y1 = p[i].y;
30             for (int j = i+1; j < n; j ++)
31             {
32                 int x2 = p[j].x;
33                 int y2 = p[j].y;
34                 int x3 = x1+(y1-y2);
35                 int y3 = y1-(x1-x2);
36                 int x4 = x2+(y1-y2);
37                 int y4 = y2-(x1-x2);
38                 if (tt.count(x3*N+y3) && (tt.count(x4*N+y4)))// 查找坐标是否存在
39                     ++ans;
40                 x3 = x1-(y1-y2);
41                 y3 = y1+(x1-x2);
42                 x4 = x2-(y1-y2);
43                 y4 = y2+(x1-x2);
44                 if (tt.count(x3*N+y3) && (tt.count(x4*N+y4)))
45                     ++ans;
46             }
47         }
48         printf("%d\n",ans/4);
49     }
50     return 0;
51 }
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posted @ 2013-08-22 10:14  N_ll  阅读(305)  评论(0)  编辑  收藏  举报