[bzoj 1468][poj 1741]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

---

又一道点分治膜版题
我果然是只会做膜版啊
还是贴上vector造图的代码吧
虽然它RE了
用数组膜你邻接表AC
我也很迷啊

好好好一个小时后的我发现自己一个小时前太sb了
居然忘了复原vector G了

 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #include<vector>
 using namespace std;

 inline int read(){
     char ch;
     int re=;
     bool flag=;
     while((ch=getchar())!='-'&&(ch<''||ch>''));
     ch=='-'?flag=:re=ch-'';
     while((ch=getchar())>=''&&ch<='')  re=re*+ch-'';
     return flag?-re:re;
 }

 struct edge{
     int to,w;
     edge(int to=,int w=):
         to(to),w(w){}
 };
 const int maxn=;
 vector<edge> G[maxn];
 int son[maxn],F[maxn];
 int d[maxn],deep[maxn];
 bool vis[maxn];
 int sum,ans,root;
 int n,K;

 inline void add_edge(int from,int to,int w){
     G[from].push_back(edge(to,w));
     G[to].push_back(edge(from,w));
 }

 bool init(){
     n=read();
     if(!n)  return ;
     memset(G,,sizeof G);
     memset(vis,,sizeof vis);
     K=read();
     for(int i=;i<n-;i++){
         int from=read(),to=read(),w=read();
         add_edge(from,to,w);
     }
     return ;
 }

 void getroot(int x,int fa){
     son[x]=;
     F[x]=;
     int dd=G[x].size();
     for(int i=;i<dd;i++){
         edge &e=G[x][i];
         if(e.to!=fa&&!vis[e.to]){
             getroot(e.to,x);
             son[x]+=son[e.to];
             F[x]=max(F[x],son[e.to]);
         }
     }
     F[x]=max(F[x],sum-son[x]);
     if(F[x]<F[root])  root=x;
 }

 void getdeep(int x,int fa){
     deep[++deep[]]=d[x];
     int dd=G[x].size();
     for(int i=;i<dd;i++){
         edge &e=G[x][i];
         if(e.to!=fa&&!vis[e.to]){
             d[e.to]=d[x]+e.w;
             getdeep(e.to,x);
         }
     }
 }

 int calc(int x,int now){
     d[x]=now;
     deep[]=;
     getdeep(x,);
     sort(deep+,deep+deep[]+);
     int t=;
     for(int l=,r=deep[];l<r;){
         if(deep[l]+deep[r]<=K){
             t+=r-l;
             l++;
         }
         else  r--;
     }
     return t;
 }

 void work(int x){
     ans+=calc(x,);
     vis[x]=;
     int dd=G[x].size();
     for(int i=;i<dd;i++){
         edge &e=G[x][i];
         if(!vis[e.to]){
             ans-=calc(e.to,e.w);
             sum=son[e.to];
             root=;
             getroot(e.to,root);
             work(root);
         }
     }
 }

 void solve(){
     sum=F[root=]=n;
     ans=;
     getroot(,);
     work(root);
     printf("%d\n",ans);
 }

 int main(){
     //freopen("temp.in","r",stdin);
     while(init())
         solve();
     return ;
 }

---

他说你任何为人称道的美丽  不及他第一次遇见你