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 10005int 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sizeint 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;}