hdu 4123 Bob’s Race 树形DP + RMQ

//	hdu 4123 Bob’s Race 树形DP + RMQ
//
//	解题思路:
//
//		先求出每个点到树中其他点的最大距离.这个可以看hdu
//	的2196 那个computer.然后用两个RMQ,一个保持最大值,一个
//	保持最小值.用所谓的尺取法,求连续的一段区间的最大值和
//	最小值.然后就可以啦~~~
//
//	感悟:
//		
//		开始训练的时候,这种dp没有写过,自己yy也没有啥好的
//	之后发现,这是经典的树形dp....其实当时我想好了后面的
//	RMQ的想法,可是并不会预处理.....哎,,,还是太年轻,继续加油
//	吧~~~fighting!!!


#include <cstring>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <vector>
#include <set>
#include <queue>
#include <map>
#define For(x,a,b,c) for (int x = a; x <= b; x += c)
#define Ffor(x,a,b,c) for (int x = a; x >= b; x -= c)
#define cls(x,a) memset(x,a,sizeof(x))
using namespace std;
typedef long long ll;

const int MAX_N = 50008;

const int INF = 0x7f7f7f7f;
const ll MOD = 1e6;
int N,M;

struct node{
	int v;
	int w;
	node(){

	}
	node(int v,int w):v(v),w(w){

	}
};

vector<node> g[MAX_N];

int d[MAX_N][3];


void dfs1(int u,int fa){
	int first = 0;
	int second = 0;

	for (int i = 0 ;i < g[u].size();i ++){
		int v = g[u][i].v;

		if (v == fa)
			continue;

		dfs1(v,u);

		int w = g[u][i].w;

		int tmp = d[v][0] + w;

		if (first <= tmp){
			second = first;
			first = tmp;
		}else if (second < tmp){
			second = tmp;
		}
	}

	d[u][0] = first;
	d[u][1] = second;
}

void dfs2(int u,int fa){


	for (int i = 0 ;i < g[u].size();i ++){
		int v = g[u][i].v;

		if (v == fa)
			continue;

		int w = g[u][i].w;

		d[v][2] = max(d[v][2],max(d[u][2],d[v][0] + w == d[u][0] ? d[u][1] : d[u][0]) + w);

		dfs2(v,u);
	}

}

int qmax[MAX_N][20];
int qmin[MAX_N][20];



void input(){

	for (int i = 1; i <= N;i ++)
		g[i].clear();

	for (int i = 1;i < N;i ++){
		int u,v,w;
		scanf("%d%d%d",&u,&v,&w);
		g[u].push_back(node(v,w));
		g[v].push_back(node(u,w));
	}
}

void init_rmq(){

	for (int i = 1;i <= N;i ++){
		for (int j = 0;j < 20 ;j ++){
			qmin[i][j] = INF;
		}
	}

	for (int i = 1;i <= N;i ++){
		qmax[i][0] = max(d[i][0],d[i][2]);
		qmin[i][0] = qmax[i][0];
	}


	for (int j = 1;(1<<j) <= N;j ++){
		for (int i = 1;i + (1 << j) <= N;i ++){
			qmax[i][j] = max(qmax[i][j-1],qmax[i + (1 << (j-1))][j-1]);
			qmin[i][j] = min(qmin[i][j-1],qmin[i + (1 << (j-1))][j-1]);
		}
	}

}

int get_max(int L,int R){
	int k = 0;

	while(1 << (k+1) <= R - L + 1) k ++;

	return max(qmax[L][k],qmax[R - (1<<k) + 1][k]);
}

int get_min(int L,int R){
	int k = 0;

	while(1 << (k + 1) <= ( R - L + 1)) k ++;

	return min(qmin[L][k],qmin[R - (1 << k) + 1][k]);
}



int get_ans(int Q){
	int L = 1,R = 1;
	int	ans = 0;
	while(L <= N && R <= N){
		while(R <= N && abs(get_max(L,R) - get_min(L,R)) <= Q){
			R++;
		}
		ans = max(ans, R - L);
		L++;
	}
	return ans;
}

void solve(){
	dfs1(1,-1);

	for (int i = 1;i <= N;i ++)
		d[i][2] = 0;

	dfs2(1,-1);

	init_rmq();
	

	for (int i = 1;i <= M;i ++){
		int Q;
		scanf("%d",&Q);

		//printf("%d\n",Q);

		printf("%d\n",get_ans(Q));
	}
}

int main(){
	//freopen("1.in","r",stdin);
	while(scanf("%d%d",&N,&M)!=EOF){
		if (!N && !M)
			break;
		input();
		solve();
	}
	return 0;
}