【POJ - 1458】Common Subsequence（动态规划）

Common Subsequence

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Descriptions:

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

题目连接：https://vjudge.net/problem/POJ-1458

题目大意

给出两个字符串，先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串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

设数组dp[i][j],i表示第一个字符串的位置i，j表示第二个字符串的位置j，如果s1[i]==s2[j]j，那么dp[i][j]=dp[i-1][j-1]+1.否则dp[i][j]=max(dp[i-1][j],dp[i][j-1]).理解不了的，可以按样例一画一个表格试试就知道了

AC代码

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#define ME0(X) memset((X), 0, sizeof((X)))
using namespace std;
string s1,s2;
int dp[][];
int main()
{
    while(cin >> s1 >> s2)
    {
        ME0(dp);
        int len1=s1.length();
        int len2=s2.length();
        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]);
                }
            }
        }
        cout << dp[len1][len2] << endl;
    }
}