HDU -- 2571 -- DP

Ah... day by day...

Today, I saw a new program group established by the public, remembering that I was at a loss when I learned that it was a family planning task last summer ..

I remember reading the video and learning C. I found it too difficult. I learned Java again and found it easier to get started with Java than C...

Forget it now =-=

I just want to say to my schoolmates: Don't rush to ask which book is suitable for getting started it's a long road

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I still like to complain. =-=,

I thought it was DFS long ago...

In fact, this question has something in common with dataworks ..

DP [I] [J] + = max (DP [I-1] [J], DP [I] [J-1], DP [I, j * k])... as for Boundary processing, this state equation is the core of this question.

At first, I wanted to remove array A and open two Arrays for ease of understanding. Then I used an array.

Every question can be optimized as much as possible if it is too big ..

 1 #include <iostream> 2 #include <algorithm> 3 using namespace std; 4  5 const int inf = -0x3f3f3f3f; 6 int maze[25][1010]; 7 int dp[25][1010]; 8  9 int main()10 {11     int t , n , m;12     while( cin >> t )13     {14         while( t-- )15         {16             cin >> n >> m;17             for( int i = 1 ; i<=n ; i++ )18             {19                 for( int j = 1 ; j<=m ; j++ )20                 {21                     cin >> maze[i][j];22                 }23             }24             for( int i = 0 ; i<=n ; i++ )25                 dp[i][0] = inf;26             for( int j = 0 ; j<=m ; j++ )27                 dp[0][j] = inf;28             dp[0][1] = dp[1][0] = 0;29             for( int i = 1 ; i<=n ; i++ )30             {31                 for( int j = 1 ; j<=m ; j++ )32                 {33                     dp[i][j] = max( dp[i-1][j] , dp[i][j-1] );34                     for( int k = 1 ; k<j ; k++ )35                     {36                         if( j%k==0 )37                             dp[i][j] = max( dp[i][j] , dp[i][k] );38                     }39                     dp[i][j] += maze[i][j];40                 }41             }42             cout << dp[n][m] << endl;43         }44     }45     return 0;46 }

View code

 

 1 #include <iostream> 2 #include <algorithm> 3 using namespace std; 4  5 const int inf = -0x3f3f3f3f; 6 int dp[25][1010]; 7  8 int main() 9 {10     int t , n , m , temp;11     while( cin >> t )12     {13         while( t-- )14         {15             cin >> n >> m;16             for( int i = 1 ; i<=n ; i++ )17             {18                 for( int j = 1 ; j<=m ; j++ )19                 {20                     cin >> dp[i][j];21                 }22             }23             for( int i = 1 ; i<=n ; i++ )24             {25                 for( int j = 1 ; j<=m ; j++ )26                 {27                     temp = inf;28                     if( i==1 && j==1 )29                         continue;30                     else if( i>=2 && j==1 )31                         temp = max( temp , dp[i-1][j] );32                     else if( i==1 && j>=2 )33                         temp = max( temp , dp[i][j-1] );34                     else35                         temp = max( dp[i-1][j] , dp[i][j-1] );36                     for( int k = 1 ; k<j ; k++ )37                     {38                         if( j%k==0 )39                             temp = max( temp , dp[i][k] );40                     }41                     dp[i][j] += temp;42                 }43             }44             cout << dp[n][m] << endl;45         }46     }47     return 0;48 }

View code

 

 

Today:

After a class reunion for a year, everyone had a tacit understanding of the hotel under the hotel =-=