【HDU】1159 Common Subsequence（DP、最长公共子序列）

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本文介绍了解决HDU 1159问题——寻找两个字符串的最长公共子序列的方法。通过动态规划算法实现，具体步骤为使用二维数组dp存储中间结果，并基于输入字符串逐字符比较，更新dp数组以找到最长公共子序列的长度。

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【HDU】1159 Common Subsequence （DP、最长公共子序列）

【题目链接】http://acm.hdu.edu.cn/showproblem.php?pid=1159

题目内容

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

题目大意

求最长公共子序列

解题思路

DP，dp[i][j]表示a[0]-a[i]和b[0]-b[j]间的最长公共子序列。
打表可知：当a[i]==b[j]时，dp[i][j]=dp[i-1][j-1]+1;
当a[i]!=b[j]时，dp[i][j]=max(dp[i-1][j],dp[i][j-1]).

AC代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<string>
#include<algorithm>
#include<math.h>
#include<limits.h>
#include<stack>
#include<queue>
#define LL long long
using namespace std;
string a,b;
int dp[1002][1002];
int main()
{
    int lena,lenb,i,j;
    while(cin>>a)
    {
        cin>>b;
        lena=a.length();
        lenb=b.length();
        memset(dp,0,sizeof(dp));
        for(i=0;i<lena;i++)
        {
            for(j=0;j<lenb;j++)
            {
                if(a[i]==b[j])
                    dp[i+1][j+1]=dp[i][j]+1;
                else dp[i+1][j+1]=max(dp[i+1][j],dp[i][j+1]);
            }
        }
        printf("%d\n",dp[lena][lenb]);
    }
    return 0;
}