hdu-acm steps Common Subsequence

/*这道题是很明显的dp题，状态方程有点不大好想，也许是我刚刚接触dp的缘故吧。dp[i][j]表示字符串s1取前i个字符s2取前j个字符时最大公共子序列的大小，这样的如果s1[i]==s2[j],dp[i][j]=d[i-1][j-1]+1;

如果s1[1]!=s2[j],dp[i][j]=max{dp[i-1][j],dp[i][j-1]};*/

#include"iostream"

#include"stdio.h"

#include"algorithm"

#include"string.h"

#include"ctype.h"

#include"cmath"

#include"queue"

#define mx 1005

#define inf 32767

#define max(a,b) a>b?a:b

using namespace std;

int dp[mx][mx];

char s1[mx],s2[mx];

int main()

{

while(scanf("%s%s",s1+1,s2+1)!=EOF)

{

int len1=strlen(s1+1);

int len2=strlen(s2+1);

int i,j;

memset(dp,0,sizeof(dp));

for(i=1;i<=len1;i++)

{

for(j=1;j<=len2;j++)

{

if(s1[i]==s2[j])

{

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

}

else

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

}

}

cout<<dp[len1][len2]<<endl;

}

return 0;

}