POJ-1458 Common Subsequence(线性动规,最长公共子序列问题)

Common Subsequence Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 44464 Accepted: 18186 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. 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 abfcab programming contest abcd mnp Sample Output

4 2 0

裸的最长公共子序列问题: 状态转移方程: if(s1[i]==s2[j]) dp[i][j]=dp[i-1][j-1]+1; else dp[i][j]=max(dp[i-1][j],dp[i][j-1]);

#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <stdlib.h>
#include <string>

using namespace std;
char s1[300];
char s2[300];
int dp[300][300];
int main()
{

    while(scanf("%s%s",&s1,&s2)!=EOF)
    {
        int len1=strlen(s1);
        int len2=strlen(s2);
                 memset(dp,0,sizeof(dp));
        for(int i=0;i<len1;i++)
        {
            for(int j=0;j<len2;j++)
            {
                if(s1[i]==s2[j])
                    dp[i+1][j+1]=dp[i][j]+1;
                else
                    dp[i+1][j+1]=max(dp[i][j+1],dp[i+1][j]);

            }
        }
        cout<<dp[len1][len2]<<endl;
    }
    return 0;
}

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