数据结构（树，点分治）：POJ 1741 Tree

 

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
　　点分治模板……

 #include <algorithm>
 #include <iostream>
 #include <cstring>
 #include <cstdio>
 using namespace std;
 const int maxn=;
 int cnt,n,k,N;
 bool vis[maxn];
 int fir[maxn],to[maxn<<],nxt[maxn<<],val[maxn<<];
 void addedge(int a,int b,int v){
     nxt[++cnt]=fir[a];fir[a]=cnt;val[cnt]=v;to[cnt]=b;
 }

 int rt,sz[maxn],son[maxn];
 int st[maxn],tot,dis[maxn];
 void Get_RT(int x,int fa){
     sz[x]=;son[x]=;
     for(int i=fir[x];i;i=nxt[i])
         if(to[i]!=fa&&!vis[to[i]]){
             Get_RT(to[i],x);
             sz[x]+=sz[to[i]];
             son[x]=max(sz[to[i]],son[x]);
         }
     son[x]=max(son[x],N-sz[x]);
     if(!rt||son[rt]>son[x])rt=x;
 }

 void DFS(int x,int fa){
     st[++tot]=dis[x];
     for(int i=fir[x];i;i=nxt[i])
         if(to[i]!=fa&&!vis[to[i]]){
             dis[to[i]]=dis[x]+val[i];
             DFS(to[i],x);
         }
 }

 int Calc(int x,int d){
     int ret=;tot=;
     dis[x]=d;DFS(x,);
     sort(st+,st+tot+);
     int l=,r=tot;
     while(l<r){
         if(st[l]+st[r]>k)r-=;
         else {ret+=r-l;l+=;}
     }
     return ret;
 }

 int Solve(int x){
     vis[x]=true;
     int ret=Calc(x,);
     for(int i=fir[x];i;i=nxt[i])
         if(!vis[to[i]]){
             ret-=Calc(to[i],val[i]);
             N=sz[to[i]];rt=;
             Get_RT(to[i],);
             ret+=Solve(rt);
         }
     return ret;
 }

 int main(){
     while(true){
         scanf("%d%d",&n,&k);
         if(!n&&!k)break;cnt=;N=n;
         memset(vis,,sizeof(vis));
         memset(fir,,sizeof(fir));
         for(int i=,a,b,v;i<n;i++){
             scanf("%d%d%d",&a,&b,&v);
             addedge(a,b,v);addedge(b,a,v);
         }
         Get_RT(,);
         printf("%d\n",Solve(rt));
     }
     return ;
 }