Farm Irrigation 

 1 #include <iostream>
 2 using namespace std;
 3 
 4 struct Node
 5 {
 6     bool t, b, l, r;
 7 }map[55][55];
 8 
 9 int p[100001];
10 
11 void make_set()
12 {
13     for(int i = 0; i < 100001; ++i)
14         p[i] = i;
15 }
16 
17 int find_set(int x)
18 {
19     if(x != p[x])
20         p[x] = find_set(p[x]);
21     return p[x];
22 }
23 
24 void union_set(int x, int y)
25 {
26     x = find_set(x);
27     y = find_set(y);
28     if(x != y)
29         p[y] = x;
30 }
31 
32 int main()
33 {
34     int m, n;
35     while(scanf("%d%d", &m, &n) && n >= 0 && m >= 0)
36     {
37         int i, j;
38         char c;
39         memset(map, 0, sizeof(map));
40         for(i = 0; i < m; ++i)
41             for(int j = 0; j < n; ++j)
42             {
43                 cin >> c;
44                 if(c == 'A' || c == 'B' || c == 'E' || c == 'H'\
45                     || c == 'G' || c == 'J' || c == 'K')
46                     map[i][j].t = true;
47                 if(c == 'C' || c == 'D' || c == 'E' || c == 'H'\
48                     || c == 'I' || c == 'J' || c == 'K')
49                     map[i][j].b = true;
50                 if(c == 'A' || c == 'C' || c == 'F' || c == 'G'\
51                     || c == 'H' || c == 'I' || c == 'K')
52                     map[i][j].l = true;
53                 if(c == 'B' || c == 'D' || c == 'F' || c == 'G'\
54                     || c == 'I' || c == 'J' || c == 'K')
55                     map[i][j].r = true;
56             }
57         make_set();    
58         for(i = 1; i < n; ++i)
59         {
60             if(map[0][i].l == true && map[0][i-1].r == true)
61                 union_set(i, i-1);/////////
62         }
63         for(i = 1; i < m; ++i)
64         {
65             if(map[i][0].t == true && map[i-1][0].b == true)
66                 union_set(i*n, (i-1)*n);/////////
67         }
68         for(i = 1; i < m; ++i)
69             for(j = 1; j < n; ++j)
70             {
71                 if(map[i][j].t && map[i-1][j].b)
72                 {
73                     union_set(i*n+j, (i-1)*n+j);
74                 }
75                 if(map[i][j].l && map[i][j-1].r)
76                 {
77                     union_set(i*n+j, i*n+j-1);
78                 }
79             }
80         int count = 0;
81         for(i = 0; i < m * n; ++i)
82             if(p[i] == i)
83                 count++;
84         cout << count << endl;
85     }
86     return 0;
87 }

　

自己思路比较清晰的一次并查集使用，就是内存有点大儿，不过思路比较简单，也很清楚

原来也还好，关键是数组开得太大了。。。

posted @ 2011-07-16 13:58  georgechen_ena  阅读(108)  评论(0)    收藏  举报