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.
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.
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：
1 program rrr(input,output); 2 type 3 etype=record 4 t,l,next:longint; 5 end; 6 var 7 e:array[0..20020]of etype; 8 a,q,father,siz,f,w:array[0..10010]of longint; 9 v:array[0..10010]of boolean; 10 n,m,i,min,x,y,l,cnt,r,h,t,ans:longint; 11 procedure add(x,y,l:longint); 12 begin 13 inc(cnt);e[cnt].t:=y;e[cnt].l:=l;e[cnt].next:=a[x];a[x]:=cnt; 14 end; 15 function max(a,b:longint):longint; 16 begin 17 if a>b then exit(a) else exit(b); 18 end; 19 procedure sort(q,h:longint); 20 var 21 i,j,x,t:longint; 22 begin 23 i:=q;j:=h;x:=w[(i+j)>>1]; 24 repeat 25 while w[i]<x do inc(i); 26 while x<w[j] do dec(j); 27 if i<=j then 28 begin 29 t:=w[i];w[i]:=w[j];w[j]:=t; 30 inc(i);dec(j); 31 end; 32 until i>j; 33 if j>q then sort(q,j); 34 if i<h then sort(i,h); 35 end; 36 function sum(h,t:longint):longint; 37 var 38 i,j,ans:longint; 39 begin 40 sort(h,t); 41 ans:=0;j:=t; 42 for i:=h to t do 43 begin 44 while (w[i]+w[j]>m) and (j>i) do dec(j); 45 if i=j then break; 46 ans:=ans+j-i; 47 end; 48 exit(ans); 49 end; 50 procedure solve(k:longint); 51 var 52 i,j:longint; 53 begin 54 h:=0;t:=1;q[1]:=k;father[k]:=0; 55 while h<t do 56 begin 57 inc(h); 58 i:=a[q[h]]; 59 while i<>0 do 60 begin 61 if not v[e[i].t] and (e[i].t<>father[q[h]]) then 62 begin 63 father[e[i].t]:=q[h]; 64 inc(t);q[t]:=e[i].t; 65 end; 66 i:=e[i].next; 67 end; 68 end; 69 if t=1 then exit; 70 for i:=1 to t do begin siz[q[i]]:=1;f[q[i]]:=0; end; 71 min:=n; 72 for i:=t downto 2 do 73 begin 74 r:=max(f[q[i]],t-siz[q[i]]); 75 if r<min then begin min:=r;j:=q[i]; end; 76 inc(siz[father[q[i]]],siz[q[i]]); 77 if siz[q[i]]>f[father[q[i]]] then f[father[q[i]]]:=siz[q[i]]; 78 end; 79 if f[k]<min then j:=k; 80 r:=a[j];cnt:=0; 81 while r<>0 do 82 begin 83 if not v[e[r].t] then 84 begin 85 h:=0;t:=1;q[1]:=e[r].t;father[e[r].t]:=j;w[cnt+1]:=e[r].l; 86 while h<t do 87 begin 88 inc(h); 89 i:=a[q[h]]; 90 while i<>0 do 91 begin 92 if not v[e[i].t] and (e[i].t<>father[q[h]]) then 93 begin 94 father[e[i].t]:=q[h]; 95 inc(t);q[t]:=e[i].t;w[cnt+t]:=w[cnt+h]+e[i].l; 96 end; 97 i:=e[i].next; 98 end; 99 end; 100 ans:=ans-sum(cnt+1,cnt+t); 101 cnt:=cnt+t; 102 end; 103 r:=e[r].next; 104 end; 105 inc(cnt);w[cnt]:=0; 106 ans:=ans+sum(1,cnt); 107 v[j]:=true;i:=a[j]; 108 while i<>0 do begin if not v[e[i].t] then solve(e[i].t);i:=e[i].next; end; 109 end; 110 begin 111 assign(input,'r.in');assign(output,'r.out');reset(input);rewrite(output); 112 while true do 113 begin 114 read(n,m);if (n=0) and (m=0) then break; 115 for i:=1 to n do a[i]:=0;cnt:=0; 116 for i:=1 to n-1 do begin readln(x,y,l);add(x,y,l);add(y,x,l); end; 117 ans:=0; 118 for i:=1 to n do v[i]:=false; 119 solve(1); 120 writeln(ans); 121 end; 122 close(input);close(output); 123 end.