POJ1741 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

LouTiancheng@POJ

 

树的重心：树上有一个结点,其所有的子树中最大的子树节点数最少,这个点就是这棵树的重心

点分治：先找出树的重心作为树根，统计其子树的答案，然后递归处理各子树，并去除重复算的答案。

 

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
const int mxn=30200;
//base
int n,k;
int ans;
//node
struct edge{
int nx,to;
int w;
}e[40020];
int head[mxn],size[mxn],mc[mxn],dis[mxn];//邻接表队首  点的子树尺寸 点重量 结点深度
int cnt;
//tree
int root,mn=mxn;
int tot;
bool vis[mxn];
//
void add_edge(int u,int v,int len){
e[++cnt]=(edge){head[u],v,len};head[u]=cnt;
e[++cnt]=(edge){head[v],u,len};head[v]=cnt;
}
void dfssize(int u,int fa){//求结点子树尺寸
size[u]=1;
mc[u]=0;
for(int i=head[u];i;i=e[i].nx){
int v=e[i].to;
if(v!=fa && !vis[v]){
dfssize(v,u);
size[u]+=size[v];
mc[u]=size[v]>mc[u]? size[v]:mc[u];
}
}
return;
}
void findmc(int u,int fa,int rt){//求树的重心
mc[u]=max(mc[u],size[rt]-size[u]);
if(mc[u]<mn){mn=mc[u];root=u;}//符合要求则更新树根
for(int i=head[u];i;i=e[i].nx){
int v=e[i].to;
if(v!=fa && !vis[v])findmc(v,u,rt);
}
return;
}
void dist(int u,int fa,int d){//求u的子树的结点深度
dis[++tot]=d;
for(int i=head[u];i;i=e[i].nx){
int v=e[i].to;
if(v!=fa && !vis[v])
dist(v,u,d+e[i].w);
}
}
int calc(int x,int d){//计算以x为中心，所有经过x的路径上符合要求的点对的数量
int res=0;
tot=0;
dist(x,0,d);//计算x为根的树的各结点深度
sort(dis+1,dis+tot+1);//深度由小到大排序
int l=1,r=tot;
while(l<r){
while(dis[l]+dis[r]>k && l<r )r--;//两结点深度相加就是距离
res+=r-l;
l++;
}
return res;
}

void dfs(int rt){//处理以rt为根的子树
//    printf("test!: %d  %d\n",rt,root);
mn=n;//记得
dfssize(rt,0);
findmc(rt,0,rt);
ans+=calc(root,0);//在以rt为根的子树中找到重心后，以其为中心计算符合要求的点对数
vis[root]=1;
for(int i=head[root];i;i=e[i].nx){
int v=e[i].to;
if(!vis[v]){
ans-=calc(v,e[i].w);//减去以v为中心的答案数，避免之后计算重复
dfs(v);
}
}
}
int main(){
while(scanf("%d",&n) && n){
memset(vis,0,sizeof(vis));
memset(head,0,sizeof(head));
ans=0;
cnt=0;//初始化！初始化！
scanf("%d",&k);
for(int i=1,u,v,d;i<n;i++){
scanf("%d%d%d",&u,&v,&d);
add_edge(u,v,d);
}
dfs(1);//随便找个结点开始计算
printf("%d\n",ans);
}
return 0;
}