poj3356 dp

 1 //Accepted    4100 KB    0 ms
 2 //类似poj1080
 3 //dp[i][j]表示s1用前i个，s2用前j个的最少匹配步数
 4 //dp[i][j]=min(dp[i][j-1]+1,dp[i-1][j-1](s1[i-1]==s2[j-1])时 or dp[i-1][j-1]+1 (s1[i-1]!=s2[j-1]时))
 5 //注意初始化条件
 6 //dp[i][0]=inf;
 7 //dp[0][i]=i;
 8 //dp[0][0]=0;
 9 #include <cstdio>
10 #include <cstring>
11 #include <iostream>
12 using namespace std;
13 const int imax_n = 1005;
14 const int inf = 100000000;
15 char s1[imax_n];
16 int n1;
17 char s2[imax_n];
18 int n2;
19 int dp[imax_n][imax_n];
20 void Dp()
21 {
22     for (int i=1;i<=n1;i++)
23     dp[i][0]=inf;
24     for (int i=1;i<=n2;i++)
25     dp[0][i]=i;
26     dp[0][0]=0;
27     for (int i=1;i<=n1;i++)
28     {
29         for (int j=1;j<=n2;j++)
30         {
31             if (s1[i-1]==s2[j-1])
32             dp[i][j]=dp[i-1][j-1];
33             else
34             dp[i][j]=dp[i-1][j-1]+1;
35             if (dp[i][j]>dp[i][j-1]+1)
36             dp[i][j]=dp[i][j-1]+1;
37         }
38     }
39     printf("%d\n",dp[n1][n2]);
40 }
41 int main()
42 {
43     while (scanf("%d%s%d%s",&n1,s1,&n2,s2)!=EOF)
44     {
45         Dp();
46     }
47     return 0;
48 }