POJ-1458 Common Subsequence （最长公共子序列）

Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 39851		Accepted: 16030

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
大致题意：求两个字符串的最长公共子序列的长度。

思路：这是一个很经典的最长公共子序列（LCS）问题。动态转移方程式如下，设有字符串X和字符串Y，dp[i,j]表示的是X的钱i个字符与Y的钱j个字符的最长公共子序列的长度。如果X[i]==Y[j]，那么这个字符与之前的LCS一定可以构成一个新的LCS；如果X[i]!=Y[j]，则分别考查dp[i-1,j]和dp[i][j-1]，选择其中的较大者为LCS。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<string>
using namespace std;
const int MAX = 500;
int dp[MAX][MAX] ={0};
int main()
{
    int len1,len2;
    string str1,str2;
    while(cin>>str1>>str2)
    {
        len1=str1.length();
        len2=str2.length();
        for(int i=1;i<=len1;i++)
        {
            for(int j=1;j<=len2;j++)
            {
                if(str1[i-1]==str2[j-1])
                {
                    dp[i][j]=dp[i-1][j-1]+1;
                }
                else
                {
                    dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
                }
            }
        }
        cout<<dp[len1][len2]<<endl;
    }
    return 0;
}