题解

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;
#define Mem(a, x) memset(a, x, sizeof(a))
#define forE(i, x, y) for(int i = head[x], y; ~i && (y = e[i].to); i = e[i].nxt)
typedef long long ll;
const int maxn = 5e4 + 5;
const int N = 16;

int n, m, a[maxn];
struct Edge
{
	int nxt, to, w;
}e[maxn << 1];
int head[maxn], ecnt = -1;
void addEdge(int x, int y, int w)
{
	e[++ecnt] = (Edge){head[x], y, w};
	head[x] = ecnt;
}

int fa[N + 2][maxn], dep[maxn], p[maxn];	//p[i]表示点i属于根的哪个儿子 
ll dis1[maxn], dis[N + 2][maxn];			//dis1[i]表示点i到根节点的时间 
void dfs1(int u, int _f)
{
	for(int i = 1; (1 << i) <= dep[u]; ++i)
	{
		fa[i][u] = fa[i - 1][fa[i - 1][u]];
		dis[i][u] = dis[i - 1][u] + dis[i - 1][fa[i - 1][u]];
	}
	forE(i, u, v)
	{
		if(v == _f) continue;
		dep[v] = dep[u] + 1;
		fa[0][v] = u, dis[0][v] = e[i].w;
		dis1[v] = dis1[u] + e[i].w;
		p[v] = p[u] ? p[u] : v;
		dfs1(v, u);
	}
}

bool vis[maxn];						//vis[i]表示这个点是否驻扎了军队
vector<int> cap;		 			//cap存放第一类节点（能跳到首都的点） 
vector<ll> res, res_son;			//res存放可调度的点的剩余时间，res_son存放需要控制的儿子的所需时间 
void jump(int x, ll mid)
{
	if(dis1[x] <= mid) cap.push_back(x);		//第一类节点 
	else
	{
		for(int i = N; i >= 0; --i)				//第二类节点 
			if(fa[i][x] && dis[i][x] <= mid) mid -= dis[i][x], x = fa[i][x];
		vis[x] = 1;				//跳不到根节点，就驻扎在最上面的节点 
	}
}
bool ned[maxn];
bool dfs2(int u, int _f)
{
	if(vis[u]) return 0;
	bool flg = 0, flg_leaf = 1;
	forE(i, u, v) if(v != _f) flg |= dfs2(v, u), flg_leaf = 0;
	return flg_leaf | flg;
}
bool check(ll mid)
{
	Mem(vis, 0), cap.clear(); res.clear(); res_son.clear(); Mem(ned, 0);
	for(int i = 1; i <= m; ++i) jump(a[i], mid);		//每个点尽量往上跳 
	forE(i, 1, v) if(dfs2(v, 1)) ned[v] = 1;			//这个孩子节点需要被占领 
	sort(cap.begin(), cap.end(), [&](int a, int b) {return mid - dis1[a] < mid - dis1[b];});
	for(int i = 0; i < (int)cap.size(); ++i)			//处理第一类节点中的特殊情况 
	{
		int x = cap[i], son = p[x];
		if(ned[son] && mid - dis1[x] < dis[0][son]) ned[son] = 0;
		else res.push_back(mid - dis1[x]);
	}
	sort(res.begin(), res.end());
	forE(i, 1, v) if(ned[v]) res_son.push_back(dis[0][v]);
	sort(res_son.begin(), res_son.end());
	for(int i = 0, j = 0; i < (int)res_son.size(); ++i)	//双指针匹配，要先排序 
	{
		while(j < (int)res.size() && res[j] < res_son[i]) ++j;
		if(j == (int)res.size()) return 0;
		++j;
	}
	return 1;
}
ll solve(ll M)					//二分 
{
	ll L = 0, R = M + 1;
	while(L < R)
	{
		ll mid = (L + R) >> 1;
		if(check(mid)) R = mid;
		else L = mid + 1;
	}
	return L == M + 1 ? -1 : L;
}

int main()
{
	Mem(head, -1), ecnt = -1;
	scanf("%d", &n);
	ll Sum = 0;
	for(int i = 1, u, v, w; i < n; ++i)
	{
		scanf("%d%d%d", &u, &v, &w);
		addEdge(u, v, w), addEdge(v, u, w);
		Sum += w;
	}
	scanf("%d", &m);
	for(int i = 1; i <= m; ++i) scanf("%d", &a[i]);
	dep[1] = 1, dfs1(1, 0);			//预处理倍增数组 
	printf("%lld\n", solve(Sum));
	return 0;
}