bzoj2783

由于路径的深度是升序的
所以我们可以考虑用前缀和的思想，用sum维护点到根路径上节点和
对于每个点x存在路径和为s即这个点到根的路径上存在y,使sum[x]-sum[y]=s
这显然是可以二分的

type node=record
       po,next:longint;
     end;

var w:array[0..100010] of node;
    q,s,a,p:array[0..100010] of longint;
    len,t,n,m,x,y,ans,i:longint;

procedure add(x,y:longint);
  begin
    inc(len);
    w[len].po:=y;
    w[len].next:=p[x];
    p[x]:=len;
  end;

function find(l,r,x:longint):boolean;
  var m:longint;
  begin
    while l<=r do
    begin
      m:=(l+r) shr 1;
      if s[q[m]]=x then exit(true);
      if s[q[m]]>x then r:=m-1 else l:=m+1;
    end;
    exit(false);
  end;

procedure dfs(x:longint);
  var i,y:longint;
  begin
    inc(t);
    q[t]:=x;
    i:=p[x];
    while i<>0 do
    begin
      y:=w[i].po;
      s[y]:=s[x]+a[y];
      if find(0,t,s[y]-m) then inc(ans);
      dfs(y);
      i:=w[i].next;
    end;
    dec(t);
  end;

begin
  readln(n,m);
  for i:=1 to n do
    read(a[i]);
  for i:=1 to n-1 do
  begin
    readln(x,y);
    add(x,y);
  end;
  s[1]:=a[1];
  q[0]:=0;
  s[0]:=0;
  dfs(1);
  writeln(ans);
end.