Common Subsequence

最新推荐文章于 2019-07-14 14:53:49 发布

原创最新推荐文章于 2019-07-14 14:53:49 发布 · 204 阅读

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本文介绍了一种使用动态规划解决最长公共子序列(LCS)问题的方法。通过两个字符串序列作为输入，该算法能找出二者间最长的相同子序列长度。文章提供了完整的C++实现代码，并解释了动态规划矩阵填充的逻辑。

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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 _{i_j} = 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LCS，典型的动态规划。

当x[i]=z[j]时，dp(i,j)=d(i-1,j-1)+1;否则，dp(i,j)=max(dp(i-1,j),dp(i,j-1))。

#include<iostream>
#include<string>
using namespace std;
int dp[10005][10005];
string x,z;
void LCS(int n,int m);
int main(){

    while(cin>>x>>z){
        LCS(x.length(),z.length());
        cout<<dp[x.length()][z.length()]<<endl;
    }
    return 0;
}
void LCS(int n,int m){
    int i,j;
    for(i=0;i<=n;i++){
        dp[i][0]=0;
    }
    for(j=0;j<=m;j++){
        dp[0][j]=0;
    }
    for(i=1;i<=n;i++){
        for(j=1;j<=m;j++){
            if(x[i-1]==z[j-1]) dp[i][j]=dp[i-1][j-1]+1;
            else dp[i][j]=dp[i][j-1]>=dp[i-1][j]?dp[i][j-1]:dp[i-1][j];
        }
    }
}