【NOIP2016练习】T3 tree （树形DP）

题意：一棵有N个结点的树，每个节点上有权值c[i]

需要选出若干结点，对于任意结点他的所有祖先都被选取且选取总个数不能超过lim

在此前提下使权值和最大

n,lim<=3000

思路：WA了1次才过真是傻

f[i,j]表示选取I为根，取J个结点的最大值

用类似背包的转移，注意dp[u,i]I不能取0，因为U必取

 var dp:array[..,..]of longint;
     c:array[..]of longint;
     head,vet,next,flag,de:array[..]of longint;
     n,lim,tot,i,j,ans,x,y:longint;

 procedure add(a,b:longint);
 begin
  inc(tot);
  next[tot]:=head[a];
  vet[tot]:=b;
  head[a]:=tot;
 end;

 function min(x,y:longint):longint;
 begin
  if x<y then exit(x);
  exit(y);
 end;

 function max(x,y:longint):longint;
 begin
  if x>y then exit(x);
  exit(y);
 end;

 procedure dfs(u:longint);
 var e,v,i,j:longint;
 begin
  flag[u]:=; de[u]:=;
  e:=head[u]; dp[u,]:=; dp[u,]:=c[u];

  e:=head[u];
  while e<> do
  begin
   v:=vet[e];
   if flag[v]= then
   begin
    dfs(v);
    de[u]:=de[u]+de[v];
    for i:=min(lim,de[u]) downto  do
     for j:= to min(lim,de[v]) do
      if i-j>= then dp[u,i]:=max(dp[u,i],dp[u,i-j]+dp[v,j]);
   end;
   e:=next[e];
  end;
  {write(u,' ');
  for i:=0 to lim do write(dp[u,i],' ');
  writeln;  }
 end;

 begin
  assign(input,'1.in'); reset(input);
  assign(output,'1.out'); rewrite(output);
  readln(n,lim);
  for i:= to n do read(c[i]);
  for i:= to n- do
  begin
   readln(x,y);
   add(x,y);
   add(y,x);
  end;
  for i:= to n do
   for j:= to lim do dp[i,j]:=-maxlongint div ;
  dfs();
  ans:=-maxlongint;
  for j:= to min(de[],lim) do ans:=max(ans,dp[,j]);
  writeln(ans);
  close(input);
  close(output);
 end.