HDU 1159：Common Subsequence（LCS模板）

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 48378    Accepted Submission(s): 22242

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

LCS模板题，直接上代码

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <limits.h>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <string>
#define ll long long
#define ms(a) memset(a,0,sizeof(a))
#define pi acos(-1.0)
#define INF 0x3f3f3f3f
const double E=exp(1);
const int maxn=1e3+10;
using namespace std;
char a[maxn],b[maxn];
int dp[maxn][maxn];
int main(int argc, char const *argv[])
{
	ios::sync_with_stdio(false);
	while(cin>>a>>b)
	{
		ms(dp);
		int la=strlen(a);
		int lb=strlen(b);
		for(int i=1;i<=la;i++)
		{
			for(int j=1;j<=lb;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]);
			}
		}
		cout<<dp[la][lb]<<endl;
	}
	return 0;
}