UVa 10192 Vacation:dp&lcs

10192-vacation

Time limit:3.000 seconds

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=114&page=show_ problem&problem=1133

Water.

Complete code:

/*0.019s*/
    
#include <bits/stdc++.h>  
using namespace std;  
    
Char a[105], b[105];  
int dp[105][105];  
    
int main ()  
{  
    int lena, LenB, I, j, cas = 0;  
    while (gets (a), a[0]!= ' # ')  
    {  
        gets (b);  
        Lena = strlen (a), LenB = strlen (b);  
        memset (DP, 0, sizeof (DP));  
        for (i = 0; i < Lena. ++i) for  
            (j = 0; j < LenB; ++j)  
            {  
                if (a[i] = = B[j]) dp[i + 1][j + 1] = Dp[i][j] + 1;  
                else Dp[i + 1][j + 1] = max (Dp[i + 1][j], Dp[i][j + 1]);  
        printf ("Case #%d:you can visit at most%d cities.\n", ++cas, Dp[lena][lenb]);  
    return 0;  
}

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