Common Subsequence

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, x ij = 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.

Input

The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.

Output

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

裸的LCS，典型的动态规划。

当x[i]=z[j]时，dp(i,j)=d(i-1,j-1)+1;否则，dp(i,j)=max(dp(i-1,j),dp(i,j-1))。

#include<iostream>
#include<string>
using namespace std;
int dp[10005][10005];
string x,z;
void LCS(int n,int m);
int main(){

while(cin>>x>>z){
LCS(x.length(),z.length());
cout<<dp[x.length()][z.length()]<<endl;
}
return 0;
}
void LCS(int n,int m){
int i,j;
for(i=0;i<=n;i++){
dp[i][0]=0;
}
for(j=0;j<=m;j++){
dp[0][j]=0;
}
for(i=1;i<=n;i++){
for(j=1;j<=m;j++){
if(x[i-1]==z[j-1]) dp[i][j]=dp[i-1][j-1]+1;
else dp[i][j]=dp[i][j-1]>=dp[i-1][j]?dp[i][j-1]:dp[i-1][j];
}
}
}