HDU 1159 Common Subsequence————LCS（最长公共子序列）模板题

最新推荐文章于 2021-05-22 22:29:28 发布

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本文介绍了一种解决最长公共子序列问题的经典算法，通过动态规划的方法，有效地找到两个字符串的最长公共子序列长度。输入为两个字符串，输出为最长公共子序列的长度。

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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.

Input

abcfbc abfcab
programming contest 
abcd mnp

Output

4
2
0

Sample Input

abcfbc abfcab
programming contest 
abcd mnp

Sample Output

4
2
0

思路：

最长公共子序列模板题

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<string>
#include<iostream>
#include<stack>
#include<queue>
#include<algorithm>
#include<map>
#include<list>
#include<vector>

#define INF 0x3f3f3f3f
#define MAXN 1003

using namespace std;

int dp[MAXN][MAXN];
int main()
{
    char a[MAXN], b[MAXN];
    while(scanf("%s%s", a, b) != EOF)
    {
        int i, j;
        int len_a = strlen(a);
        int len_b = strlen(b);
        memset(dp, 0, sizeof(dp));
        for(i = 1; i <= len_a; i++)
        {
            for(j = 1; j <= len_b; j++)
            {
                if(a[i-1] == b[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[len_a][len_b]);
    }

    return 0;
}