POJ 1741 Tree 树+点分治

树的点分治

可以看09年漆子超论文,说的很详细.

Tree

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 12650		Accepted: 4025

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

/* ***********************************************
Author        :CKboss
Created Time  :2015年04月24日 星期五 09时37分44秒
File Name     :POJ1741_2.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

const int INF=0x3f3f3f3f;
const int maxn=10100;

struct Node { int to,len; };

int n,K;
vector<Node> G[maxn];
bool done[maxn];

int size,root;
int s[maxn],f[maxn],d[maxn];
vector<int> dep;
int ans;

void findRoot(int u,int fa)
{
	s[u]=1; f[u]=0;
	int rest=0;
	for(int i=0,sz=G[u].size();i<sz;i++)
	{
		int v=G[u][i].to;
		if(v==fa||done[v]) continue;
		findRoot(v,u);
		rest+=s[v];
		f[u]=max(f[u],s[v]);
	}
	s[u]+=rest;
	f[u]=max(f[u],size-s[u]);
	if(f[u]<f[root]) root=u;
}

void getDeep(int u,int fa)
{
	dep.push_back(d[u]);
	s[u]=1;
	for(int i=0,sz=G[u].size();i<sz;i++)
	{
		int v=G[u][i].to;
		if(v==fa||done[v]) continue;
		int len=G[u][i].len;
		d[v]=d[u]+len;
		getDeep(v,u);
		s[u]+=s[v];
	}
}

int calc(int u,int init)
{
	dep.clear(); d[u]=init;
	getDeep(u,-1);
	sort(dep.begin(),dep.end());
	int ret=0;
	for(int l=0,r=dep.size()-1;l<r;)
	{
		if(dep[l]+dep[r]<=K) ret+=r-l++;
		else r--;
	}
	return ret;
}

void solve(int u)
{
	ans+=calc(u,0);
	done[u]=true;
	for(int i=0,sz=G[u].size();i<sz;i++)
	{
		int v=G[u][i].to;
		if(done[v]) continue;
		int len=G[u][i].len;
		ans-=calc(v,len);
		// 递归子树
		size=s[v]; root=0; findRoot(v,u);
		solve(root);
	}
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

	f[0]=INF;

	while(scanf("%d%d",&n,&K)!=EOF)
	{
		if(n==0&&K==0) break;

		memset(done,false,sizeof(done));
		for(int i=0;i<=n+10;i++) G[i].clear();

		for(int i=0;i<n-1;i++)
		{
			int a,b,c;
			scanf("%d%d%d",&a,&b,&c);
			G[a].push_back((Node){b,c});
			G[b].push_back((Node){a,c});
		}

		size=n; ans=0;
		findRoot(1,-1);
		solve(root);
		printf("%d\n",ans);
	}
    
    return 0;
}