poj-1458-Common Subsequence-lcs-最长公共子序列-java

Common Subsequence

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 55700 Accepted: 23186

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         abfcabprogramming    contest abcd           mnp

Sample Output

420解题思路：本题裸的最长公共子序列问题 经典dp            动态转移方程                    1：char(a) == char(b) dp[i][j] = dp[i-1][j-1] +1;                                      2：char(a) != char(b) dp[i][j] = max(dp[i-1][j],dp[i][j-1],dp[i-1][j-1])

在这里为新手提个醒 子序列可以不是连续的 子串才是必须连续的 我懵了好长时间。。。。。。源码：

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        while (scanner.hasNext()) {
            String a = scanner.next();
            String b = scanner.next();
            int [][] dp = new int [a.length()+1][b.length()+1];
            for (int i = 1; i <= a.length(); i++) {
                for (int j = 1; j <= b.length(); j++) {
                    if (a.charAt(i-1) == b.charAt(j-1)) {
                        dp[i][j] = dp[i-1][j-1] +1;
                    }else {
                        dp[i][j] = Math.max(Math.max(dp[i-1][j], dp[i][j-1]),dp[i-1][j-1]);
                    }
                }
            }
            System.out.println(dp[a.length()][b.length()]);
        }
    }

}