2783: [JLOI2012]树 

 1 var
 2   max,u,v,ans,s,i,j,k,n,m:longint;
 3   p,sum:array[0..100000,0..20]of longint;
 4   f,d,x:Array[0..100000]of longint;
 5 procedure dou(a:longint);
 6   var now,i,j,k:longint;
 7   begin
 8   now:=x[a];
 9   if a=1 then begin if now=s then inc(ans); exit;end;
10   k:=trunc(ln(d[a])/ln(2));
11   for i:=k downto 0 do
12     begin
13     if (now+sum[a,i]<=s) then
14       begin
15       now:=now+sum[a,i];
16       a:=p[a,i];
17       end;
18     if now=s then break;
19     end;
20   if now=s then inc(ans);
21   end;
22 
23 begin
24   readln(n,s);
25   for i:=1 to n do
26     read(x[i]);
27   for i:=1 to n-1 do
28     begin
29     readln(u,v);
30     f[v]:=u;
31     d[v]:=d[u]+1;
32     if d[v]>max then
33       max:=d[v];
34     end;
35   k:=trunc(ln(max)/ln(2));
36   for i:=1 to n do
37     sum[i,0]:=x[f[i]];
38   for i:=1 to n do p[i,0]:=f[i];
39   for j:=1 to k do
40     for i:=1 to n do
41       begin
42       p[i,j]:=p[p[i,j-1],j-1];
43       sum[i,j]:=sum[i,j-1]+sum[p[i,j-1],j-1];
44       end;
45   for i:=1 to n do
46     if (x[i]<=s) then
47       dou(i);
48   writeln(ans);
49   end.

 

posted @ 2013-03-01 18:59  lbz007  阅读(185)  评论(0)  编辑  收藏  举报
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