Common Subsequence(dp入门题)

最新推荐文章于 2020-08-22 20:52:42 发布

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最新推荐文章于 2020-08-22 20:52:42 发布

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分类专栏： ACM

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本文介绍了一道经典的动态规划题目——求两个字符串的最长公共子序列问题，并提供了详细的解题思路及C++实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

dp给我的感觉真的很奇妙，以前刚学dfs的时候是定义这个函数有什么功能，现在是定义一个数组有什么功能。

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29393 Accepted Submission(s): 13215

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

Source

Southeastern Europe 2003

水的差不多了，开始专题了

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
char s1[1010],s2[1010];
int dp[1010][1010];
int main()
{
    while(scanf("%s%s",s1,s2)==2){
        int len1=strlen(s1);
        int len2=strlen(s2);
        for(int i=0;i<=len1;i++)
            dp[i][0]=0;
        for(int i=0;i<len2;i++)
            dp[0][i]=0;
        for(int i=1;i<=len1;i++)
            for(int j=1;j<=len2;j++)
                if(s1[i-1]==s2[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\n",dp[len1][len2]);
    }
    return 0;
}

AC之路，我选择坚持~~