Common Subsequence（LCS最长公共子序列）

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本文介绍了一个经典的计算机科学问题——最长公共子序列(LCS)问题，并提供了一段C++实现代码，该算法通过动态规划方法高效地求解两个字符串的最长公共子序列长度。

Common Subsequence

Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%lld & %llu

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, x ij = 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

题目大意：
求两字符串的最长公共子序列问题（即LCS问题）。

//AC
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[30][30];
char s1[100],s2[100];

int main()
{
    while (scanf("%s %s",s1,s2)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        for (int i=0;i<strlen(s1);i++)
        {
            for (int j=0;j<strlen(s2);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]);
            }
        }
//        for (int i=0;i<strlen(s1);i++)
//        {
//            for (int j=0;j<strlen(s2);j++)
//            {
//                printf("%d ",dp[i+1][j+1]);
//            }
//            printf("\n");
//        }
        printf("%d\n",dp[strlen(s1)][strlen(s2)]);
    }
    return 0;
}

测试数据：abcd becd

测试结果：3

图解： s1\s2 b e c d

a 0 0 0 0

b 1 1 1 1

c 1 1 2 2

d 1 1 1 3