HDU 1159 Common Subsequence 动态规划

题目链接：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): 16338    Accepted Submission(s): 6810

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

 

解题思路：动态规划方程dp[i][j] = max ( dp[i-1][k] ) (0 <= k < j) + 1;

解题代码：G++

 

// File Name: Common Subsequence 1159.cpp
// Author: sheng
// Created Time: 2013年05月14日 星期二 15时48分11秒

#include <cstdio>
#include <string.h>
#include <iostream>
using namespace std;

const int max_n = 1010;
char str1[max_n], str2[max_n];
int dp[2][max_n]; //滚动数组dp

int main()
{

    int Max;
    while (~scanf ("%s %s", str1, str2))
    {
        int len1 = strlen (str1);
        int len2 = strlen (str2);
        memset (dp, 0, sizeof (dp));
        int pos = 1; //滚动数组初始使用
        for (int i = 0; i < len1; i ++)
        {
            for (int j = 0; j < len2; j ++)
            {
                if (j == 0 || i == 0)
                    Max = 0;
                else
                {
                    Max = max (dp[pos^1][j-1], Max);
                } 
                if (str1[i] == str2[j])
                    dp[pos][j] = Max + 1;
                else dp[pos][j] = dp[pos^1][j];
            }
            pos ^= 1;//交换数组
        }
        pos ^= 1;
        for (int i = 0; i < len2; i ++)
        {
            Max = max (dp[pos][i], Max);
        }
        printf ("%d\n", Max);
    }
    return 0;
}