POJ 3267 The Cow Lexicon(dp)

题目链接

状态转移算是比较好推把，但是匹配字符串很容易会超时，跟以前某个题目的用法很类似，从第i个字母往后找假如到第j个字母匹配了一个单词,这样dp[j] = dp[i-1]+j-i-l[k]+1，取小即可。

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 using namespace std;

 7 int dp[501];

 8 char str[601][31],s1[601],l[601];

 9 int num,n,len;

10 int main()

11 {

12     int i,j,k,u;

13     scanf("%d%d",&n,&len);

14     scanf("%s",s1);

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

16     {

17         scanf("%s",str[i]);

18         l[i] = strlen(str[i]);

19     }

20     for(i = 0; i <= len-1; i ++)

21     {

22         dp[i] = i+1;

23     }

24     for(i = 0; i <= len-1; i ++)

25     {

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

27         {

28             if(s1[i] == str[j][0])

29             {

30                 num = 1;

31                 if(l[j] == 1)

32                 {

33                     dp[i] = min(dp[i],dp[i-1]);

34                     for(u = i+1;u <= len-1;u ++)

35                     {

36                         if(i-1 >= 0)

37                         dp[u] = min(dp[u],dp[i-1]+u-i-l[j]+1);

38                         else

39                         dp[u] = min(dp[u],u-i-l[j]+1);

40                     }

41                 }

42                 else

43                 {

44                     for(k = i+1; k <= len-1; k ++)

45                     {

46                         if(str[j][num] == s1[k])

47                         {

48                             num ++;

49                         }

50                         if(num == l[j])

51                         {

52                             for(u = k;u <= len-1;u ++)

53                             {

54                                 if(i-1 >= 0)

55                                 dp[u] = min(dp[u],dp[i-1]+u-i-l[j]+1);

56                                 else

57                                 dp[u] = min(dp[u],u-i-l[j]+1);

58                             }

59                             break;

60                         }

61                     }

62                 }

63             }

64         }

65     }

66     printf("%d\n",dp[len-1]);

67     return 0;

68 }