最长公共子序列

Common Subsequence

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 51421 Accepted: 21197

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         abfcabprogramming    contest abcd           mnp

Sample Output

420

 

令maxlen[i][0],maxlen[0][j]都等于零

递推公式

if(str1[i-1]==str2[j-1])
                maxlen[i][j]=maxlen[i-1][j-1]+1;
            else
                maxlen[i][j]=max(maxlen[i-1][j],maxlen[i][j-1]);

注：当str1[i-1]！=str2[j-1]时，maxlen[str1][str2]不会比maxlen[str1-1][str2]，maxlen[str1][str2-1两者之间任意一个小，也不会比两个大。

#include <iostream>#include <cstring>using namespace std;char str1[1000];char str2[1000];int maxlen[1000][1000];int main(){    while(cin >>str1>>str2)    {        int strlen1=strlen(str1);        int strlen2=strlen(str2);        for(int i=0;i<=strlen1;i++)            maxlen[i][0]=0;        for(int j=0;j<=strlen2;j++)            maxlen[0][j]=0;        for(int i=1;i<=strlen1;i++)            for(int j=1;j<=strlen2;j++)        {            if(str1[i-1]==str2[j-1])                maxlen[i][j]=maxlen[i-1][j-1]+1;            else                maxlen[i][j]=max(maxlen[i-1][j],maxlen[i][j-1]);        }        cout<<maxlen[strlen1][strlen2]<<endl;    }    return 0;}

演示图：