LCS简单问题

问题 B: 选拔赛专用题(2)

时间限制: 1 Sec  内存限制: 128 MB
提交: 52  解决: 14
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题目描述

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.

输入

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.

输出

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.

样例输入

abcfbc         abfcabprogramming    contest abcd           mnp

样例输出

420

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
 
char s1[1000],s2[1000];
int dp[1000][1000];
int len1,len2;
 
void LCS()
{
    int i,j;
    memset(dp,0,sizeof(dp));
    for(i = 1; i<=len1; i++)
    {
        for(j = 1; j<=len2; j++)
        {
            if(s1[i-1] == s2[j-1])
                dp[i][j] = dp[i-1][j-1]+1;
            else
                dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
        }
    }
}
int main()
{
    while(cin>>s1>>s2)
    {
        len1 = strlen(s1);
        len2 = strlen(s2);
        LCS();
        cout<<dp[len1][len2]<<endl;
    }
    return 0;
}