hdu1159 最大公共子串 java

最新推荐文章于 2021-03-06 02:43:12 发布

一枚软的硬币

最新推荐文章于 2021-03-06 02:43:12 发布

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分类专栏： hdu 算法随笔

本文链接：https://blog.youkuaiyun.com/q1248807225/article/details/88608505

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本文详细介绍了解决最长公共子序列问题的一种动态规划算法。通过具体实例，文章展示了如何使用二维数组来存储两个字符串之间的最大匹配长度，最终找出这两个字符串的最长公共子序列的长度。代码实现清晰，便于理解。

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http://acm.hdu.edu.cn/showproblem.php?pid=1159

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 52105 Accepted Submission(s): 24031

Problem 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, xij = 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.
The program input is from a text file. Each data set in the file contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct. 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

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 < dp.length; i++) {
                for (int j = 1; j < dp[i].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(dp[i][j-1], dp[i-1][j]);
                    }
                }
            }
            System.out.println(dp[a.length()][b.length()]);
            
        }
    }
}