poj 1741Tree

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
Source

树点分治例题

大神的论文：《分治算法在树的路径问题中的应用》

(分治思想）:把问题分成形式相同，规模更小的子问题。

对于一棵有根树， 树中满足要求的一条路径，必然是以下两种情况之一：

1、经过根节点

2、不经过根节点，在根节点的一棵子树中

对于情况2，可以递归（重复步骤1，2）求解，主要考虑情况1（因题而异，注意去重）。

找根节点就是重点了，因为要尽可能减少深度，所以要找一个使子树中size最大的尽量小（也就是 重心}）。然后使重心作为根，继续操作（记得标记是否找过）。

ps：与“最近点对”有异曲同工之妙

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int num,t,K,ans;
#define N 10005
int to[N+N],head[N],Next[N+N],val[N+N];
int a[N],deep[N],size[N];
bool flag[N];
void add(int u,int v,int z){
  ++num;
  to[num]=v;
  val[num]=z;
  Next[num]=head[u];
  head[u]=num;
}
void getsize(int u,int fa){
  size[u]=1;
  for (int i=head[u];i;i=Next[i])
  if (to[i]!=fa&&flag[to[i]]){
    getsize(to[i],u);
    size[u]+=size[to[i]];
  }
}//预处理size
int getroot(int u,int fa,int ma){
  for (int i=head[u];i;i=Next[i])
  if (to[i]!=fa&&flag[to[i]]&&size[to[i]]+size[to[i]]>ma) return getroot(to[i],u,ma);
  return u;
}//找重心
void getdeep(int u,int fa){
  a[t++]=deep[u];
  for (int i=head[u];i;i=Next[i])
  if (to[i]!=fa&&flag[to[i]]) deep[to[i]]=deep[u]+val[i],getdeep(to[i],u);
}
int calc(int u,int ss){
  t=0;
  deep[u]=ss;
  getdeep(u,-1);
  sort(a,a+t);
  int l=0,r=t-1,xx=0;
  while (l<r){
    if (a[l]+a[r]>K){--r;continue;}
    xx+=r-l;
    ++l;
  }
  return xx; 
}
void work(int u){
  flag[u]=false;
  getsize(u,-1);
  ans+=calc(u,0);
  for (int i=head[u];i;i=Next[i])
  if (flag[to[i]]){
    ans-=calc(to[i],val[i]);
    int rt=getroot(to[i],u,size[to[i]]);
    work(rt);
  }
}
int main(){
  int n;
  while (scanf("%d%d",&n,&K)!=EOF){
    if (n==0&&K==0) break;
    ans=0;
    num=0;
    for (int i=1;i<=n;++i) head[i]=0,flag[i]=true;
    for (int i=1;i<n;++i){
      int u,v,z;
      scanf("%d%d%d",&u,&v,&z);
      add(u,v,z);
      add(v,u,z);
    }
    getsize(1,-1);
    int rt=getroot(1,-1,size[1]);
    work(rt);
    printf("%d\n",ans);
  }
  return 0;
}