POJ 1458 Common Subsequence

L - Common Subsequence

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

Submit Status Practice 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

最长公共子序列问题。动态转移方程和图解如下：

转移方程：

图解：

轻松写出代码：

/*************************************************************************
	> File Name: Common_Subsequence.cpp
	> Author: Zhanghaoran
	> Mail: chilumanxi@xiyoulinux.org
	> Created Time: Sat 24 Oct 2015 10:39:26 PM CST
 ************************************************************************/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>

using namespace std;

int dp[201][201];
char str1[201];
char str2[201];
int main(void){
    while(~scanf("%s%s", str1, str2)){
        dp[0][0] = 0;
        for(int i = 1; i <= strlen(str1); i ++){
            for(int j = 1; j <= strlen(str2); 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]);
                }
            }
        }

        printf("%d
", dp[strlen(str1)][strlen(str2)]);
    }
    return 0;
}