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 [0..101,0..3] of longint;
    a:array [0..101,0..4,0..10] of longint;
begin
    readln(n,m,k);
    for i:=1 to n do for j:=1 to m do read(b[i,j]);
    if m=1 then
        begin
            for i:=0 to k do
                begin
                    a[0,1,i]:=-maxlongint div 10;
                    a[0,0,i]:=-maxlongint div 10;
                end;
            a[0,0,0]:=0;
            for i:=1 to n do
                for j:=0 to k do
                    begin
                        if a[i-1,1,j]>a[i-1,0,j] then a[i,0,j]:=a[i-1,1,j] else a[i,0,j]:=a[i-1,0,j];
                        if j=0 then
                           a[i,1,j]:=-maxlongint div 10
                        else
                            begin
                                a[i,1,j]:=a[i-1,1,j]+b[i,1];
                                if a[i,1,j]<a[i-1,0,j-1]+b[i,1] then a[i,1,j]:=a[i-1,0,j-1]+b[i,1];
                                if (j>0) and (a[i,1,j]<a[i-1,1,j-1]+b[i,1]) then a[i,1,j]:=a[i-1,1,j-1]+b[i,1];
                            end;
                    end;
            if a[n,0,k]>a[n,1,k] then writeln(a[n,0,k]) else writeln(a[n,1,k]);
            halt;
        end;
    for i:=0 to k do for j:=0 to 4 do a[0,j,i]:=-maxlongint div 10;
    a[0,0,0]:=0;
    for i:=1 to n do
        for j:=0 to k do
            begin
                for ii:=0 to 4 do
                    case ii of
                         0:begin
                                a[i,0,j]:=a[i-1,0,j];
                                for jj:=1 to 4 do if a[i,0,j]<a[i-1,jj,j] then a[i,0,j]:=a[i-1,jj,j];
                         end;
                         1..2:begin
                                   a[i,ii,j]:=a[i-1,ii,j]+b[i,ii];
                                   for jj:=0 to 4 do if (j>0) and (a[i,ii,j]<a[i-1,jj,j-1]+b[i,ii]) then a[i,ii,j]:=a[i-1,jj,j-1]+b[i,ii];
                                   if a[i,ii,j]<a[i-1,3,j]+b[i,ii] then a[i,ii,j]:=a[i-1,3,j]+b[i,ii];
                         end;
                         3:begin
                                a[i,ii,j]:=a[i-1,ii,j]+b[i,2]+b[i,1];
                                for jj:=0 to 4 do if (j>=2) and (a[i,ii,j]<a[i-1,jj,j-2]+b[i,2]+b[i,1]) then a[i,ii,j]:=a[i-1,jj,j-2]+b[i,2]+b[i,1];
                                for jj:=1 to 3 do if (j>=1) and (a[i,ii,j]<a[i-1,jj,j-1]+b[i,2]+b[i,1]) then a[i,ii,j]:=a[i-1,jj,j-1]+b[i,2]+b[i,1];
                         end;
                         4:begin
                                a[i,ii,j]:=a[i-1,ii,j]+b[i,2]+b[i,1];
                                for jj:=0 to 4 do if (j>=1) and (a[i,ii,j]<a[i-1,jj,j-1]+b[i,2]+b[i,1]) then a[i,ii,j]:=a[i-1,jj,j-1]+b[i,2]+b[i,1];
                         end;
                    end;
            end;
    tot:=-maxlongint div 10;
    for i:=0 to 4 do if tot<a[n,i,k] then tot:=a[n,i,k];
    writeln(tot);
end.