每日AC--POJ 1458 Common Subsequence -DP

Common Subsequence

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 54754 Accepted: 22811

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

import java.util.Scanner;/** * 类说明 *  * <pre> * Modify Information: * Author        Date          Description * ============ =========== ============================ * DELL          2017年8月2日    Create this file * </pre> *  */public class Main {     /**     * 最长公共子序列问题， 给定 两个字符串 求 这两个字符串的公共子序列     * 确认这个公式很重要.     *      * 动态规划问题：     *  dp[i+1][j+1] = {     *                   dp[i][j] +1,  dp[i][j+1], dp[j+1][i]  //     *                   dp[i+1][j] ,dp[i][j+1]     *                   }     */    public int getLongestCommonString(String str1, String str2){                int[][]  dp = new int[str1.length()+1][str2.length()+1];        char[] strChar1 = str1.toCharArray();        char[] strChar2 = str2.toCharArray();               for(int i = 0; i < strChar1.length; i++){            for(int j = 0; j < strChar2.length; j++){                if(strChar1[i] == strChar2[j]){                    dp[i+1][j+1] =  dp[i][j] +1;                }else{                    dp[i+1][j+1] = Math.max(dp[i+1][j], dp[i][j+1]);                }            }        }        return dp[str1.length()][str2.length()];    }        /**     * @param args     */    public static void main(String[] args) {        // TODO Auto-generated method stub        String str1 = "addadbs";        String str2 ="hdgsfvab";                String str3 = "abcfbc";        String str4 ="abfcab";                Scanner cin = new Scanner(System.in);        String str5 = "";        String str6 = "";                while(cin.hasNext()){            str5 = cin.next();            str6 = cin.next();            int ans = new Main().getLongestCommonString(str5, str6);            System.out.println(ans);        }             }}