hdu 1159 Common Subsequence

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1159

题目描述：

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. 

Input

abcfbc abfcab
programming contest
abcd mnp

Output

4
2
0

题目大意：

给你两个字符串，叫你找出最长公共子序列的长度

题目分析：

看到这种有两个字符串的问题，第一反应就是二维dp，dp[i][j]表示字符串s1到i以及字符串s2到j这一段最长公共子序列的长度

状态转移方程：

        if(s1[i]==s2[j])

                    dp[i][j]=dp[i-1][j-1]+1;

       else

                   dp[i][j]=max(dp[i-1][j-1],max(dp[i-1][j],dp[i][j-1]));

AC代码：

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iomanip>
#include<algorithm>
using namespace std;

char s1[1005],s2[1005];
int dp[1005][1005];
int l1,l2;

int main()
{
while(scanf("%s%s",s1+1,s2+1)!=EOF)
{
l1=strlen(s1+1);
l2=strlen(s2+1);
memset(dp,0,sizeof(dp));
for(int i=1;i<=l1;i++)
{
for(int j=1;j<=l2;j++)
{
if(s1[i]==s2[j])
dp[i][j]=dp[i-1][j-1]+1;
else
dp[i][j]=max(dp[i-1][j-1],max(dp[i-1][j],dp[i][j-1]));
}
}
cout<<dp[l1][l2]<<endl;
}
return 0;
}