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