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[..] of node;
     q,s,a,p:array[..] 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 ;
       if s[q[m]]=x then exit(true);
       if s[q[m]]>x then r:=m- else l:=m+;
     end;
     exit(false);
   end;

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

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