（POJ - 1458）Common Subsequence

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.

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

题目大意：给出两个字符串，要求最长公共子序列（不一定连续）。
思路：典型的dp。设f[i][j]表示第一个字符串前i个字符与第二个字符串前j个字符所能得到的最长公共子序列（s1，s2为输出的两个字符串）那么易得
**f[i][j]=f[i-1][j-1] (s1[i]==s2[j]);
f[i][j]==max(f[i-1][j],f[i][j-1]) (s1[i]!=s2[j])。**
(其中初值为：f[i][0]=0(0<=i<=len1)&&f[0][i]=0(0<=i<=len2))
显然答案便是f[len1][len2] (len1为字符串s1的长，len2为字符串s2的长)

#include
#include
#include
using namespace std;

const int maxn=10005;
char s1[maxn],s2[maxn];
int f[maxn][maxn];

int main()
{
    while(scanf("%s%s",s1+1,s2+1)!=EOF)
    {
        int len1=strlen(s1+1);
        int len2=strlen(s2+1);
        for(int i=0;i<=len1;i++) f[i][0]=0;
        for(int i=0;i<=len2;i++) f[0][i]=0;
        for(int i=1;i<=len1;i++)
            for(int j=1;j<=len2;j++)
            {
                if(s1[i]==s2[j]) f[i][j]=f[i-1][j-1]+1;
                else f[i][j]=max(f[i-1][j],f[i][j-1]);   
            }
        printf("%d\n",f[len1][len2]);
    }
    return 0;
}