poj 1458

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

420

題意：

我們稱序列 Z = < z1, z2, ..., zk >是序列X = < x1, x2, ..., xm >的子序列當且僅當存在嚴格上升的序列< i1, i2, ..., ik >，使得對j = 1, 2, ... ,k, 有xij = zj。比如Z = < a, b, f, c > 是X = < a, b,c, f, b, c >的子序列。現在給出兩個序列X 和Y，你的任務是找到X 和Y 的最大公共子序列，也就是說要找到一個最長的序列Z，使得Z 既是X 的子序列也是Y 的子序列.

思路：

設d(i,j)爲A1,A2,...Ai和B1,B2,..Bj的LCS的長度，則當A[i]=A[j]時d(I,j)d(i-1,j-1)+1,否則d(i,j)=max{d(i-1,j),d(i,j-1)},時間複雜度爲O（nm），其中n和m分別是序列A和B的長度。

代碼：

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n,m;
int const maxn=1000;
char s[maxn],t[maxn];
int dp[maxn+1][maxn+1];
void solve()
{
    for(int i=0; i<n; i++)
    {
        for(int j=0; j<m; j++)
        {
            if(s[i]==t[j])
            {
                dp[i+1][j+1]=dp[i][j]+1;
            }
            else
            {
                dp[i+1][j+1]=max(dp[i][j+1],dp[i+1][j]);

            }
        }
    }
    printf("%d\n",dp[n][m]);
}
int main()
{

    while(scanf("%s%s",&s,&t)!=EOF)
    {
        n=strlen(s);
        m=strlen(t);
        solve();
    }
    return 0;

}