POJ 1129 Channel Allocation 四色定理dfs

题目： http://poj.org/problem?id=1129

开始没读懂题，看discuss的做法，都是循环枚举的，很麻烦。然后我就决定dfs，调试了半天终于0ms A了。

 1 #include <stdio.h>
 2 #include <string.h>
 3 bool graph[26][26], vis[26][4];
 4 int n, ans;
 5 
 6 void calc()
 7 {
 8     int cnt = 0;
 9     for(int i = 0; i < 4; i++)
10     {
11         for(int j = 0; j < n; j++)
12         {
13             if(vis[j][i])
14             {
15                 cnt++;
16                 break;
17             }
18         }
19     }
20     if(cnt < ans)ans = cnt;
21 }
22 
23 void dfs(int x)
24 {
25     if(x >= n)
26     {
27         calc();
28         return;
29     }
30 
31     for(int i = 0; i < 4; i++)
32     {
33         bool ok = 1;
34         for(int j = 0; j < n; j++)
35         {
36             if((graph[j][x] && vis[j][i]) || (graph[x][j] && vis[j][i]))
37             {
38                 ok = 0;
39                 break;
40             }
41         }
42         if(ok)
43         {
44             vis[x][i] = 1;
45             dfs(x+1);
46             vis[x][i] = 0;
47         }
48     }
49 }
50 
51 int main()
52 {
53     char s[30];
54     while(scanf("%d", &n) != EOF && n)
55     {
56         ans = 0x3f3f3f3f;
57         memset(graph, 0, sizeof(graph));
58         memset(vis, 0, sizeof(vis));
59         for(int i = 0; i < n; i++)
60         {
61             scanf("%s", s);
62             for(int j = 2; s[j]; j++)
63             {
64                 graph[i][s[j]-'A'] = 1;
65             }
66         }
67         vis[0][0] = 1;
68         dfs(1);
69         if(ans == 1)
70             printf("1 channel needed.\n");
71         else printf("%d channels needed.\n", ans);
72     }
73     return 0;
74 }
View Code

 

posted @ 2013-08-20 09:44  Anti-Magic  阅读(262)  评论(0)  编辑  收藏  举报