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。

 #include <stdio.h>
 #include <string.h>
 bool graph[][], vis[][];
 int n, ans;

 void calc()
 {
     int cnt = ;
     for(int i = ; i < ; i++)
     {
         for(int j = ; j < n; j++)
         {
             if(vis[j][i])
             {
                 cnt++;
                 break;
             }
         }
     }
     if(cnt < ans)ans = cnt;
 }

 void dfs(int x)
 {
     if(x >= n)
     {
         calc();
         return;
     }

     for(int i = ; i < ; i++)
     {
         bool ok = ;
         for(int j = ; j < n; j++)
         {
             if((graph[j][x] && vis[j][i]) || (graph[x][j] && vis[j][i]))
             {
                 ok = ;
                 break;
             }
         }
         if(ok)
         {
             vis[x][i] = ;
             dfs(x+);
             vis[x][i] = ;
         }
     }
 }

 int main()
 {
     char s[];
     while(scanf("%d", &n) != EOF && n)
     {
         ans = 0x3f3f3f3f;
         memset(graph, , sizeof(graph));
         memset(vis, , sizeof(vis));
         for(int i = ; i < n; i++)
         {
             scanf("%s", s);
             for(int j = ; s[j]; j++)
             {
                 graph[i][s[j]-'A'] = ;
             }
         }
         vis[][] = ;
         dfs();
         if(ans == )
             printf("1 channel needed.\n");
         else printf("%d channels needed.\n", ans);
     }
     return ;
 }