Common Subsequence HDU

 Common Subsequence

 HDU - 1159 

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. 

Input

abcfbc abfcabprogramming contest abcd mnp

Output

420

Sample Input

abcfbc abfcabprogramming contest abcd mnp

Sample Output

420

AC代码：

#include<stdio.h>#include<iostream>#include<string.h>#include<algorithm>#include<cmath>#define maxn 505using namespace std;int main(){int len1,len2,i,j,flag,dp[maxn][maxn];char s[maxn],t[maxn];while(scanf("%s%s",s,t)!=EOF){memset(dp,0,sizeof(dp));len1=strlen(s);len2=strlen(t);for(i=0;i<len1;++i){for(j=0;j<len2;++j){if(s[i]==t[j]){dp[i+1][j+1]=dp[i][j]+1;}else{dp[i+1][j+1]=max(dp[i][j+1],dp[i+1][j]);}}}/*for(i=0;i<=len1;++i){              //dp数组的值 for(j=0;j<=len2;++j){printf("%d ",dp[i][j]);}cout<<endl;}*/cout<<dp[len1][len2]<<endl;}return 0;}