HDU 1159 Common Subsequence (动态规划最长公共子序列)

Posted dyhaohaoxuexi

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了HDU 1159 Common Subsequence (动态规划最长公共子序列)相关的知识,希望对你有一定的参考价值。

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 55301    Accepted Submission(s): 25537

 

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

题目大意与分析

就是求最长公共子序列,动态规划即可。

dp[i][j]代表A序列前i-1个元素与B序列前j-1个元素的最长公共子序列的个数,两层循环,相等就++,否则取当前最大的

代码

#include<bits/stdc++.h>

using namespace std;

string x,y;
int i,j,dp[1005][1005];

int main()

    while(cin>>x>>y)
    
        memset(dp,0,sizeof(dp));
        for(i=0;i<x.size();i++)
        
            for(j=0;j<y.size();j++)
            
                if(x[i]==y[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]);
            
        
        cout<<dp[x.size()][y.size()]<<endl;
    

 

 

以上是关于HDU 1159 Common Subsequence (动态规划最长公共子序列)的主要内容,如果未能解决你的问题,请参考以下文章

HDU 1159 Common Subsequence

HDU 1159 Common Subsequence(裸LCS)

hdu 1159 Common Subsequence(最长公共子序列)

hdu-1159 Common Subsequence

题解报告:hdu 1159 Common Subsequence

hdu 1159 Common Subsequence(lcs)