POJ 1458 Common Subsequence （动态规划）

题目传送门 POJ 1458

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：
　　给出两个字符串，求出这样的一 个最长的公共子序列的长度：子序列 中的每个字符都能在两个原串中找到， 而且每个字符的先后顺序和原串中的 先后顺序一致。

解题思路：
　　用一个数组dp[i][j] 存  到s1的第i位为止 s2的前j位中有dp[i][j]个字符的先后顺序和原串中的先后顺序一致。
　　状态转移方程为：
　　　　如果s1[i]和s2[j]相等 则 dp[i+1][j+1]=dp[i][j]+1
　　　　　                 否则 dp[i+1][j+1]=max(dp[i+1][j],dp[i][j+1])
　　结果输出dp[len(s1)][len(s2)]即可。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = ;
char s1[N],s2[N];
int dp[N][N];
int main()
{
    while(~scanf("%s%s",s1,s2))
    {
        int len1=strlen(s1),len2=strlen(s2);
        memset(dp,,sizeof(dp));
        for (int i=; i<len1; i++)
            for (int j=; j<len2; j++)
            {
                if (s1[i]==s2[j])
                    dp[i+][j+]=dp[i][j]+;
                else
                    dp[i+][j+]=max(dp[i+][j],dp[i][j+]);
            }
        printf("%d\n",dp[len1][len2]);
    }
    return ;
}