Bzoj1084: [scoi2005] maximum submatrix

1084: [scoi2005] maximum submatrix time limit: 10 sec memory limit: 162 MB
Submit: 1129 solved: 578
[Submit] [Status] Description

Here is a matrix of N * M. Please select K sub-matrices to maximize the total score of K sub-matrices. Note: The selected K submatrices cannot overlap with each other.

Input

First behavior n, m, K (1 ≤ n ≤, 1 ≤ m ≤ 2, 1 ≤ k ≤ 10 ), the following n lines describe the values of each element in each row of the matrix (the absolute value of each element's score cannot exceed 32767 ).

Output

Only one row is the maximum value of the score of K sub-matrix.

Sample input3 2 2
1-3
2 3
-2 3
Sample output9
Hint Source

Question:

Difficulties of DP:

1. It is the most important to use several-dimensional variables to represent the complete information of a State.

2. Status transfer enumeration: it is relatively easy to know what status the status may be transferred to or what status may be transferred.

If the status transfer is complicated, it may be that the status representation is too simple.

If you cannot think of a good method, you need to take the status into details.

For this question

F [I, j, k] indicates that I is obtained from the left column, J is obtained from the right, K is obtained.

Of course it is not all, or the decision on the left is I, and the decision on the right is J. The K decision should be made.

The transfer of status is very obvious.

Code:

 1 uses math; 2 const maxn=100+10;maxk=15; 3 var g:array[0..maxn,0..maxk] of longint; 4     s:array[1..2,0..maxn] of longint; 5     f:array[0..maxn,0..maxn,0..maxk] of longint; 6     i,j,k,l,t,n,m,x:longint; 7 procedure init; 8  begin 9    readln(n,m,t);10    for i:=1 to n do11     begin12       for j:=1 to m do begin read(x);s[j,i]:=s[j,i-1]+x;end;13       readln;14     end;15  end;16 procedure work1;17  begin18   fillchar(g,sizeof(g),0);19   for k:=1 to t do20    for i:=k to n do21     begin22     g[i,k]:=g[i-1,k];23     for j:=k-1 to i-1 do24      g[i,k]:=max(g[i,k],g[j,k-1]+s[1,i]-s[1,j]);25     end;26   writeln(g[n,t]);27  end;28 procedure work2;29  begin30   fillchar(f,sizeof(f),0);31   for k:=1 to t do32    for i:=0 to n do33     for j:=0 to n do34      begin35        if (i=0) and (j=0) then continue;36        f[i,j,k]:=max(f[i-1,j,k],f[i,j-1,k]);37        for l:=0 to j-1 do38         f[i,j,k]:=max(f[i,j,k],f[i,l,k-1]+s[2,j]-s[2,l]);39        for l:=0 to i-1 do40         f[i,j,k]:=max(f[i,j,k],f[l,j,k-1]+s[1,i]-s[1,l]);41        if i=j then42         for l:=0 to i-1 do43          f[i,j,k]:=max(f[i,j,k],f[l,l,k-1]+s[1,i]-s[1,l]+s[2,j]-s[2,l]);44      end;45   writeln(f[n,n,t]);46  end;47 48 begin49   assign(input,‘input.txt‘);assign(output,‘output.txt‘);50   reset(input);rewrite(output);51   init;52   if m=1 then work1 else work2;53   close(input);close(output);54 end.   

View code