POJ1458 Common Subsequence

文章鏈接http://poj.org/problem?id=1458

Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 48657		Accepted: 20064

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         abfcab
programming    contest 
abcd           mnp

Sample Output

4
2
0

Source

Southeastern Europe 2003

這道題是水體，可以作爲熟練代碼用的LCS，除了輸入要處理一下，別的沒了

代碼如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cctype>
#define MAXN 720
using namespace std;

int main()
{
    string bck;
    while(getline(cin,bck))
    {
        string s,t;
        int dp[MAXN][MAXN];
        memset(dp,0,sizeof(dp));
        int ct = 0,len_bck = bck.length();
        while(isalpha(bck[ct]) && ct<len_bck) s += bck[ct++];
        while(!isalpha(bck[ct]) && ct<len_bck) ct++;
        while(isalpha(bck[ct]) && ct<len_bck) t += bck[ct++];

        int len_s = s.length(),len_t = t.length();
        for(int i=0;i<len_s;i++)
        {
            for(int j=0;j<len_t;j++)
            {
                if(s[i] == t[j]) dp[i+1][j+1] = dp[i][j] + 1;
                else dp[i+1][j+1] = max(dp[i+1][j],dp[i][j+1]);
            }
        }
        printf("%d\n",dp[len_s][len_t]);
    }
    return 0;
}