tzoj1510 Common Subsequence（最长公共子序列LCS模板）

时间限制(普通/Java):1000MS/10000MS     内存限制:65536KByte

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 a text file. Each data set in the file 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 abfcab
programming contest
abcd mnp

 

样例输出

4
2
0

 

题目来源

Southeastern Europe 2003

 

 

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 1000;
char a[N],b[N];
int dp[N][N];
int main()
{
    int lena,lenb,i,j;
    while(scanf("%s%s",a,b)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        lena=strlen(a);
        lenb=strlen(b);
        for(i=1;i<=lena;i++)
        {
            for(j=1;j<=lenb;j++)
            {
                if(a[i-1]==b[j-1])
                {
                    dp[i][j]=dp[i-1][j-1]+1;
                }
                else
                {
                    dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
                }
            }
        }
        printf("%d\n",dp[lena][lenb]);
    }
    return 0;
}