依旧LCS，dp解决，套模板即可。
注意字符串中可以有空格。
解题代码如下：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
#define max(a,b) ((a)>(b)?(a):(b))
#define maxn 105
char Seq1[maxn], Seq2[maxn];
int N1, N2;
int LongLen[maxn][maxn];
int dp(int i, int j)
{
if(i==-1 || j==-1) return 0;
if(LongLen[i][j]>=0) return LongLen[i][j];
if(Seq1[i]==Seq2[j]) return LongLen[i][j] = dp(i-1,j-1)+1;
else return LongLen[i][j] = max( dp(i-1,j), dp(i,j-1));
}
int main()
{
int Case = 0;
while(gets(Seq1) && Seq1[0]!='#')
{
gets(Seq2);
N1 = strlen(Seq1); N2 = strlen(Seq2);
memset(LongLen,-1,sizeof(LongLen));
printf("Case #%d: you can visit at most %d cities.
", ++Case, dp(N1-1,N2-1));
}
return 0;
}