bzoj1296

首先先预处理每行刷1~m次最多能正确涂出多少格

然后把每行涂色看做一个物品，当重量为j（这行涂了j次），价值为对应能正确涂出的格子数；

总重量为k，然后做分组背包即可

var f:array[0..60,0..60,0..60] of longint;
    sum:array[0..60,0..60] of longint;
    dp:array[0..60,0..3000] of longint;
    t,x,p,n,m,k,l,i,j:longint;
    s:string;

function max(a,b:longint):longint;
  begin
    if a>b then exit(a) else exit(b);
  end;

begin
  readln(n,m,t);
  for i:=1 to n do
  begin
    readln(s);
    l:=length(s);
    for j:=1 to l do
    begin
      x:=ord(s[j])-48;
      sum[i,j]:=sum[i,j-1]+x;
    end;
  end;
  for i:=1 to n do
  begin
    for j:=1 to m do
    begin
      f[i,j,0]:=0;
      for k:=1 to j do
      begin
        for l:=j-1 downto 0 do
        begin
          p:=sum[i,j]-sum[i,l];
          f[i,j,k]:=max(f[i,j,k],f[i,l,k-1]+max(p,j-l-p));
        end;
      end;
    end;
  end;
  for i:=1 to n do
  begin
    for j:=t downto 0 do
    begin
      for k:=0 to m do
        if j-k>=0 then dp[i,j]:=max(dp[i,j],dp[i-1,j-k]+f[i,m,k])
        else break;
    end;
  end;
  writeln(dp[i,t]);
end.