1084: [SCOI2005]最大子矩阵

1084: [SCOI2005]最大子矩阵

Time Limit: 10 Sec  Memory Limit: 162 MB
Submit: 1325  Solved: 670
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Description

这里有一个n*m的矩阵，请你选出其中k个子矩阵，使得这个k个子矩阵分值之和最大。注意：选出的k个子矩阵不能相互重叠。

Input

第一行为n,m,k（1≤n≤100,1≤m≤2,1≤k≤10），接下来n行描述矩阵每行中的每个元素的分值(每个元素的分值的绝对值不超过32767)。

Output

只有一行为k个子矩阵分值之和最大为多少。

Sample Input

3 2 2
1 -3
2 3
-2 3

Sample Output

9

HINT

 

Source

 题解：看了半天不知道怎么办才好，直到发现M<=2（HansBug:呵呵你早说就俩列得了呗呵呵哒）假如这样的话那么直接根据此行俩数的不同情况讨论下不就得啦。。。情况繁一点

 program bzoj1084;

 {$mode objfpc}{$H+}

 uses
   Classes, SysUtils
   { you can add units after this };

 var
     i,j,k,l,m,n,tot,ii,jj:longint;
     b:array [..,..] of longint;
     a:array [..,..,..] of longint;
 begin
     readln(n,m,k);
     for i:= to n do for j:= to m do read(b[i,j]);
     if m= then
         begin
             for i:= to k do
                 begin
                     a[,,i]:=-maxlongint div ;
                     a[,,i]:=-maxlongint div ;
                 end;
             a[,,]:=;
             for i:= to n do
                 for j:= to k do
                     begin
                         if a[i-,,j]>a[i-,,j] then a[i,,j]:=a[i-,,j] else a[i,,j]:=a[i-,,j];
                         if j= then
                            a[i,,j]:=-maxlongint div
                         else
                             begin
                                 a[i,,j]:=a[i-,,j]+b[i,];
                                 if a[i,,j]<a[i-,,j-]+b[i,] then a[i,,j]:=a[i-,,j-]+b[i,];
                                 if (j>) and (a[i,,j]<a[i-,,j-]+b[i,]) then a[i,,j]:=a[i-,,j-]+b[i,];
                             end;
                     end;
             if a[n,,k]>a[n,,k] then writeln(a[n,,k]) else writeln(a[n,,k]);
             halt;
         end;
     for i:= to k do for j:= to  do a[,j,i]:=-maxlongint div ;
     a[,,]:=;
     for i:= to n do
         for j:= to k do
             begin
                 for ii:= to  do
                     case ii of
                          :begin
                                 a[i,,j]:=a[i-,,j];
                                 for jj:= to  do if a[i,,j]<a[i-,jj,j] then a[i,,j]:=a[i-,jj,j];
                          end;
                          ..:begin
                                    a[i,ii,j]:=a[i-,ii,j]+b[i,ii];
                                    for jj:= to  do if (j>) and (a[i,ii,j]<a[i-,jj,j-]+b[i,ii]) then a[i,ii,j]:=a[i-,jj,j-]+b[i,ii];
                                    if a[i,ii,j]<a[i-,,j]+b[i,ii] then a[i,ii,j]:=a[i-,,j]+b[i,ii];
                          end;
                          :begin
                                 a[i,ii,j]:=a[i-,ii,j]+b[i,]+b[i,];
                                 for jj:= to  do if (j>=) and (a[i,ii,j]<a[i-,jj,j-]+b[i,]+b[i,]) then a[i,ii,j]:=a[i-,jj,j-]+b[i,]+b[i,];
                                 for jj:= to  do if (j>=) and (a[i,ii,j]<a[i-,jj,j-]+b[i,]+b[i,]) then a[i,ii,j]:=a[i-,jj,j-]+b[i,]+b[i,];
                          end;
                          :begin
                                 a[i,ii,j]:=a[i-,ii,j]+b[i,]+b[i,];
                                 for jj:= to  do if (j>=) and (a[i,ii,j]<a[i-,jj,j-]+b[i,]+b[i,]) then a[i,ii,j]:=a[i-,jj,j-]+b[i,]+b[i,];
                          end;
                     end;
             end;
     tot:=-maxlongint div ;
     for i:= to  do if tot<a[n,i,k] then tot:=a[n,i,k];
     writeln(tot);
 end.