POJ-1458 Common Subsequence ( DP )

题目链接: http://acm.hust.edu.cn/vjudge/contest/view.action?cid=68966#problem/L

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 找两个串的最大公共子串
状态转移方程 以dp[i][j]表示串1前i个字符和串2前j个字符的最大公共子串，则当str1[i]==dp[j]时有dp[i][j] = dp[i - 1][j - 1] + 1，否则dp[i][j] = max( dp[i - 1][j], dp[i][j - 1] ),此结论可用反正证明

#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

int dp[1005][1005];

int main(){
    ios::sync_with_stdio( false );

    string str1, str2;
    while( cin >> str1 >> str2 ){
        memset( dp, 0, sizeof( dp ) );

        for( int i = 0; i < str1.size(); i++ )
            for( int j = 0; j < str2.size(); j++ ){
                if( str1[i] == str2[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] );
            }

        cout << dp[str1.size()][str2.size()] << endl;
    }

    return 0;
}