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 }

 

posted @ 2012-12-04 20:56  Naix_x  阅读(207)  评论(0)  编辑  收藏  举报