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 }

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