Common Subsequence [dp]

最长公共子序列求解

最新推荐文章于 2019-10-15 20:37:02 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动态规划方法。通过两个字符串作为输入，算法能够找出这两个字符串间的最长公共子序列长度。示例代码展示了如何使用二维数组存储中间结果以避免重复计算，实现高效的解决方案。

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

题解

dp入门题

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAX_N 500
using namespace std;
int dp[2][MAX_N];
char a[MAX_N],b[MAX_N];

int main()
{
    while(~scanf("%s%s",a+1,b+1)){
        int A=strlen(a+1);
        int B=strlen(b+1);
        memset(dp,0,sizeof(dp));
        for(int j=1;j<=A;j++)
            for(int k=1;k<=B;k++){
                if(a[j]==b[k])
                    dp[j&1][k]=dp[(j-1)&1][k-1]+1;
                else
                    dp[j&1][k]=max(dp[(j-1)&1][k],dp[j&1][k-1]);
            }
        printf("%d\n",dp[A&1][B]);
    }
    return 0;
}