【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[1..3000,0..3000]of longint;
    c:array[1..3000]of longint;
    head,vet,next,flag,de:array[1..10000]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]:=1; de[u]:=1;
 e:=head[u]; dp[u,0]:=0; dp[u,1]:=c[u];

 e:=head[u];
 while e<>0 do
 begin
  v:=vet[e];
  if flag[v]=0 then
  begin
   dfs(v);
   de[u]:=de[u]+de[v];
   for i:=min(lim,de[u]) downto 2 do
    for j:=0 to min(lim,de[v]) do
     if i-j>=1 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:=1 to n do read(c[i]);
 for i:=1 to n-1 do
 begin
  readln(x,y);
  add(x,y);
  add(y,x);
 end;
 for i:=1 to n do
  for j:=0 to lim do dp[i,j]:=-maxlongint div 3;
 dfs(1);
 ans:=-maxlongint;
 for j:=0 to min(de[1],lim) do ans:=max(ans,dp[1,j]);
 writeln(ans);
 close(input);
 close(output);
end.