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Common Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38425 Accepted Submission(s): 17634

Problem Description A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = <x1, x2, ..., xm> another sequence Z = <z1, z2, ..., zk> is a subsequence of X if there exists a strictly increasing sequence <i1, i2, ..., ik> of indices of X such that for all j = 1,2,...,k, xij = zj. For example, Z = <a, b, f, c> is a subsequence of X = <a, b, c, f, b, c> with index sequence <1, 2, 4, 6>. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y.
The program input is from a text file. Each data set in the file contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct. For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line.

Sample Input abcfbc abfcab programming contest abcd mnp

Sample Output 4 2 0

Source Southeastern Europe 2003 本題就是經典的lcs問題哈，還是老套路，把一個問題分解為若幹小問題。 得到如下遞推式： 下面附代碼：

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<cmath>
 4 #include<algorithm>
 5 using namespace
 
 std;
 6 const int Max = 1111;
 7 char st1[Max],st2[Max];
 8 int dp[Max][Max];
 9 int main()
10 {
11     while(~scanf("%s %s",st1,st2))
12     {
13         memset(dp,0,sizeof(dp));
14         int m=strlen(st1);
15         int n=strlen(st2);
16         for(int i=0;i<m;i++)
17         for(int k=0;k<n;k++)
 
18         {
19             if(st1[i]==st2[k])
20             dp[i+1][k+1]=dp[i][k]+1;
21             else
22             dp[i+1][k+1]=max(dp[i+1][k],dp[i][k+1]);
23         }
24         printf("%d\n",dp[m][n]);
25     }
26     return 0;
27 }

hdu-1159 Common Subsequence （dp中的lcs問題）