Common Subsequence（LCS最长公共子序列）

Common Subsequence

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%lld & %llu

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


题目大意：
    求两字符串的最长公共子序列问题（即LCS问题）。

//AC#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int dp[30][30];char s1[100],s2[100];int main(){    while (scanf("%s %s",s1,s2)!=EOF)    {        memset(dp,0,sizeof(dp));        for (int i=0;i<strlen(s1);i++)        {            for (int j=0;j<strlen(s2);j++)            {                if (s1[i]==s2[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 (int i=0;i<strlen(s1);i++)//        {//            for (int j=0;j<strlen(s2);j++)//            {//                printf("%d ",dp[i+1][j+1]);//            }//            printf("\n");//        }        printf("%d\n",dp[strlen(s1)][strlen(s2)]);    }    return 0;}

测试数据：abcd becd

测试结果：3

图解：    s1\s2 b  e  c  d

                  a   0  0  0  0  

            b   1  1  1  1

                  c   1  1  2  2

                  d   1  1  1  3