poj1741 Tree（点分治）

Description

Give a tree with n vertices,each edge has a length(positive integer less than 1001). 
Define dist(u,v)=The min distance between node u and v. 
Give an integer k,for every pair (u,v) of vertices is called valid if and only if dist(u,v) not exceed k. 
Write a program that will count how many pairs which are valid for a given tree. 

Input

The input contains several test cases. The first line of each test case contains two integers n, k. (n<=10000) The following n-1 lines each contains three integers u,v,l, which means there is an edge between node u and v of length l. 
The last test case is followed by two zeros. 

Output

For each test case output the answer on a single line.

Sample Input

5 4
1 2 3
1 3 1
1 4 2
3 5 1
0 0

Sample Output

8


题目大意：

给定一棵N个结点的带权树。 定义dist(u，v)=u，v两点间的路径长度，路径的长度定义为路径上所有边的权和。 给定一个K，如果对于不同的两个结点a，b，如果满足dist(a，b)≤K，则称(a，b)为合法点对。 求合法点对个数。

这道题是点分治的模板题，具体可参考2009年漆子超的国家集训队论文《分治算法在树的路径问题中的应用》
第一次写点分治，非常不熟练，也犯了一些变量打错的sb错误。
AC代码：

program rrr(input,output);
type
  etype=record
     t,l,next:longint;
  end;
var
  e:array[0..20020]of etype;
  a,q,father,siz,f,w:array[0..10010]of longint;
  v:array[0..10010]of boolean;
  n,m,i,min,x,y,l,cnt,r,h,t,ans:longint;
procedure add(x,y,l:longint);
begin
   inc(cnt);e[cnt].t:=y;e[cnt].l:=l;e[cnt].next:=a[x];a[x]:=cnt;
end;
function max(a,b:longint):longint;
begin
   if a>b then exit(a) else exit(b);
end;
procedure sort(q,h:longint);
var
  i,j,x,t:longint;
begin
   i:=q;j:=h;x:=w[(i+j)>>1];
   repeat
     while w[i]<x do inc(i);
     while x<w[j] do dec(j);
     if i<=j then
        begin
           t:=w[i];w[i]:=w[j];w[j]:=t;
           inc(i);dec(j);
        end;
   until i>j;
   if j>q then sort(q,j);
   if i<h then sort(i,h);
end;
function sum(h,t:longint):longint;
var
  i,j,ans:longint;
begin
   sort(h,t);
   ans:=0;j:=t;
   for i:=h to t do
      begin
         while (w[i]+w[j]>m) and (j>i) do dec(j);
         if i=j then break;
         ans:=ans+j-i;
      end;
   exit(ans);
end;
procedure solve(k:longint);
var
  i,j:longint;
begin
   h:=0;t:=1;q[1]:=k;father[k]:=0;
   while h<t do
      begin
         inc(h);
         i:=a[q[h]];
         while i<>0 do
            begin
               if not v[e[i].t] and (e[i].t<>father[q[h]]) then
                  begin
                     father[e[i].t]:=q[h];
                     inc(t);q[t]:=e[i].t;
                  end;
               i:=e[i].next;
            end;
      end;
   if t=1 then exit;
   for i:=1 to t do begin siz[q[i]]:=1;f[q[i]]:=0; end;
   min:=n;
   for i:=t downto 2 do
      begin
         r:=max(f[q[i]],t-siz[q[i]]);
         if r<min then begin min:=r;j:=q[i]; end;
         inc(siz[father[q[i]]],siz[q[i]]);
         if siz[q[i]]>f[father[q[i]]] then f[father[q[i]]]:=siz[q[i]];
      end;
   if f[k]<min then j:=k;
   r:=a[j];cnt:=0;
   while r<>0 do
      begin
         if not v[e[r].t] then
            begin
               h:=0;t:=1;q[1]:=e[r].t;father[e[r].t]:=j;w[cnt+1]:=e[r].l;
               while h<t do
                  begin
                     inc(h);
                     i:=a[q[h]];
                     while i<>0 do
                        begin
                           if not v[e[i].t] and (e[i].t<>father[q[h]]) then
                              begin
                                 father[e[i].t]:=q[h];
                                 inc(t);q[t]:=e[i].t;w[cnt+t]:=w[cnt+h]+e[i].l;
                              end;
                           i:=e[i].next;
                        end;
                  end;
               ans:=ans-sum(cnt+1,cnt+t);
               cnt:=cnt+t;
            end;
         r:=e[r].next;
      end;
   inc(cnt);w[cnt]:=0;
   ans:=ans+sum(1,cnt);
   v[j]:=true;i:=a[j];
   while i<>0 do begin if not v[e[i].t] then solve(e[i].t);i:=e[i].next; end;
end;
begin
   assign(input,'r.in');assign(output,'r.out');reset(input);rewrite(output);
   while true do
      begin
         read(n,m);if (n=0) and (m=0) then break;
         for i:=1 to n do a[i]:=0;cnt:=0;
         for i:=1 to n-1 do begin readln(x,y,l);add(x,y,l);add(y,x,l); end;
         ans:=0;
         for i:=1 to n do v[i]:=false;
         solve(1);
         writeln(ans);
      end;
   close(input);close(output);
end.