POJ 2817 WordStack(状态压缩DP)

题目链接

以前的坑，以前做的时候，用暴力各种跪，其实是状态压缩DP。暴力处理出来后，状态+标记最后一个单词。

 1 #include <cstdio>

 2 #include <cstring>

 3 #include <cmath>

 4 #include <iostream>

 5 using namespace std;

 6 char p[12][12];

 7 int o[12][12];

 8 int dp[1026][12];

 9 int dfs(int x,int y)

10 {

11     int i,j,k,len1,len2,ret,ans;

12     len1 = strlen(p[x]);

13     len2 = strlen(p[y]);

14     ans = 0;

15     for(i = 0; i <= len1-1; i ++)

16     {

17         ret = 0;

18         for(j = 0,k = i; j <= len2-1&&k <= len1-1; j ++,k ++)

19         {

20             if(p[x][k] == p[y][j])

21                 ret ++;

22         }

23         if(ans < ret)

24             ans = ret;

25     }

26     for(i = 0; i <= len2-1; i ++)

27     {

28         ret = 0;

29         for(j = 0,k = i; j <= len1-1&&k <= len2-1; j ++,k ++)

30         {

31             if(p[y][k] == p[x][j])

32                 ret ++;

33         }

34         if(ans < ret)

35             ans = ret;

36     }

37     return ans;

38 }

39 int main()

40 {

41     int i,j,num,temp,n,u,k;

42     while(scanf("%d%*c",&n)!=EOF)

43     {

44         if(!n) break;

45         memset(dp,-1,sizeof(dp));

46         for(i = 0; i < n; i ++)

47             scanf("%s",p[i]);

48         for(i = 0; i < n-1; i ++)

49         {

50             for(j = i+1; j < n; j ++)

51             {

52                 o[i][j] = o[j][i] = dfs(i,j);

53             }

54         }

55         for(i = 0; i < n; i ++)

56         {

57             dp[1<<i][i] = 0;

58         }

59         num = n;

60         for(i = 1; i < n; i ++)

61         {

62             temp = 0;

63             for(j = 1; j < 1<<n; j ++)

64             {

65                 for(u = 0; u < n; u ++)

66                 {

67                     for(k = 0; k < n; k ++)

68                     {

69                         if(j&(1<<k)) continue;

70                         if(dp[j][u] == -1) continue;

71                         if(dp[j+(1<<k)][k] < dp[j][u] + o[u][k])

72                         dp[j+(1<<k)][k] = dp[j][u] + o[u][k];

73                     }

74                 }

75             }

76         }

77         int maxz = 0;

78         for(i = 0;i < n;i ++)

79         maxz = max(dp[(1<<n)-1][i],maxz);

80         printf("%d\n",maxz);

81     }

82     return 0;

83 }