【USACO】Apple Catching 2004,Nov

时间：近1h

总结：速度实在是太慢了……主要是问题没想清楚

方程：记f[i,j]为考虑前i个果子，剩余移动次数还有j次时能吃到的最大果子数

则　　

　　 f[i-1,j]+1 (t[i]=t[i-1])

f[i,j]= max{f[i-1,j+1]+1,f[i-1,k](the nearest k such that t[k]=t[i])} (t[i]<>t[i-1])

边界：f[1,0..k-1]=1 f[1,k]=1(t[1]=1) f[1,k]=0(t[1]=2)

var
   n,k,i,j,max,q:integer;
   t:array[1..1000] of integer;
   f:array[0..1000,0..30] of integer;

begin

   assign(input,'bcatch.in');
   reset(input);
   assign(output,'bcatch.out');
   rewrite(output);
   readln(n,k);
   for i:=1 to n do readln(t[i]);
   for i:=0 to k do f[1,i]:=1;
   if t[1]=2 then f[1,k]:=0;
   for i:=2 to n do
   for j:=k downto 0 do
   begin
      if (j=k)and(t[i]=2) then begin
         f[i,j]:=f[i-1,j];
         end
      else begin
         if t[i]=t[i-1] then f[i,j]:=f[i-1,j]+1 else f[i,j]:=f[i-1,j+1]+1;
         for q:=i-1 downto 1 do if t[q]=t[i] then break;
         if f[q,j]+1>f[i,j] then f[i,j]:=f[q,j]+1;
        if f[i,j]>max then max:=f[i,j];
      end;
   end;

   writeln(max);

   close(input);
   close(output);
end.

转载于:https://www.cnblogs.com/OmegaIota/p/3332002.html