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Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34400    Accepted Submission(s): 15711


Problem 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, 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. 
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. 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 


思路：
一道典型的求最长公共子序列的题，dp[i][j]代表以a的第i个元素和b的第j个元素结尾的两个串的公共子序列，编列所有dp[i][j]，如果a[i]==b[j]说明这一个元素可以加进去，就等于dp[i-1][j-1]+1，如果不等就在 a向前一位 和 b向前一位 中找最长的。水题。


代码： 
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

char a[1000+5],b[1000+5];
int dp[1000+5][1000+5];

int main()
{
	//freopen("in.txt","r",stdin);
	while(scanf("%s%s",a,b)!=EOF)
	{
		int str1=strlen(a);
		int str2=strlen(b);
		memset(dp,0,sizeof(dp));
		for(int i=1;i<=str1;i++)
		{
			for(int j=1;j<=str2;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[str1][str2]);
	}
	return 0;
}