「2017 山东一轮集训 Day5」距离

/*
写完开店再写这个题目顿时神清气爽， 腰也不疼了， 眼也不花了

首先考虑将询问拆开， 就是查询一些到根的链和点k的关系

根据我们开店的结论， 一个点集到一个定点的距离和可以分三部分算
那么就很简单了吧QAQ， 在树上可持久化弄一下 

*/

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#define ll long long
#define mmp make_pair
#define M 200010
using namespace std;
int read()
{
	int nm = 0, f = 1;
	char c = getchar();
	for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
	return nm * f;
}

int type, n, q, sz[M], son[M], fa[M], top[M], dfn[M], ver[M], dft, cnt, p[M], deep[M];
ll dis[M], sum[M], w[M], ans;
vector<pair<int,int> > to[M];
int lc[20000100], rc[20000100], v[20000100], rt[M];
ll t[20000100];

void dfs(int now, int f)
{
	deep[now] = deep[f] + 1;
	sz[now] = 1;
	fa[now] = f;
	for(int i = 0; i < to[now].size(); i++)
	{
		int vj = to[now][i].first, v = to[now][i].second;
		if(vj == f) continue;
		dis[vj] = dis[now] + v;
		ver[vj] = v;
//		deep[vj] = deep[now] + 1;
		dfs(vj, now);
		if(sz[vj] > sz[son[now]]) son[now] = vj;
		sz[now] += sz[vj];
	}
}

void dfs(int now)
{
	dfn[now] = ++cnt;
	w[cnt] = ver[now];
	if(son[now])
	{
		top[son[now]] = top[now];
		dfs(son[now]);
	}
	for(int i = 0; i < to[now].size(); i++)
	{
		int vj = to[now][i].first;
		if(vj == fa[now] || vj == son[now]) continue;
		top[vj] = vj;
		dfs(vj);
	}
}

void modify(int last, int &now, int l, int r, int ln, int rn)
{
	now = ++cnt;
	lc[now] = lc[last], rc[now] = rc[last], t[now] = t[last], v[now] = v[last];
	if(l == ln && r == rn)
	{
		v[now]++;
		return;
	}
	t[now] += w[rn] - w[ln - 1];
	int mid = (l + r) >> 1;
	if(ln > mid) modify(rc[last], rc[now], mid + 1, r, ln, rn);
	else if(rn <= mid) modify(lc[last], lc[now], l, mid, ln, rn);
	else modify(lc[last], lc[now], l, mid, ln, mid), modify(rc[last], rc[now], mid + 1, r, mid + 1, rn);
}

ll query(int now, int l, int r, int ln, int rn)
{
	ll ans = 1ll * (w[rn] - w[ln - 1]) * v[now];
	if(l == ln && r <= rn) return ans + t[now];
	int mid = (l + r) >> 1;
	if(rn <= mid) return ans + query(lc[now], l, mid, ln, rn);
	else if(ln > mid) return ans + query(rc[now], mid + 1, r, ln, rn);
	else return ans + query(lc[now], l, mid, ln, mid) + query(rc[now], mid + 1, r, mid + 1, rn);
}

void Modify(int &now, int x)
{
	for(; top[x] != 1; x = fa[top[x]]) modify(now, now, 1, n, dfn[top[x]], dfn[x]);
	modify(now, now, 1, n, dfn[1], dfn[x]);
}

void work(int now)
{
	rt[now] = rt[fa[now]];
	Modify(rt[now], p[now]);
	sum[now] = dis[p[now]] + sum[fa[now]];
	for(int i = 0; i < to[now].size(); i ++)
	{
		int vj = to[now][i].first;
		if(vj == fa[now]) continue;
		work(vj);
	}
}

ll Query(int x, int k)
{
	if(k == 0) return 0;
//	cout << k << " ";
	ll ans = sum[x];
	ans += dis[k] * deep[x];
	for(; top[k] != 1; k = fa[top[k]]) ans -= 2ll * query(rt[x], 1, n, dfn[top[k]], dfn[k]);
	ans -= 2ll *query(rt[x], 1, n, dfn[1], dfn[k]);
//	cout << x  << " " << ans << "!!!!!!!\n";
	return ans;
}

int LCA(int a, int b)
{
	while(top[a] != top[b])
	{
		if(deep[top[a]] < deep[top[b]]) swap(a, b);
		a = fa[top[a]];
	}
	if(deep[a] > deep[b]) swap(a, b);
	return a;
}

int main()
{
	type = read();
	n = read(), q = read();
	for(int i = 1; i < n; i++)
	{
		int vi = read(), vj = read(), v = read();
		to[vi].push_back(mmp(vj, v));
		to[vj].push_back(mmp(vi, v));
	}
	for(int i = 1; i <= n; i++) p[i] = read();
	dfs(1, 0);
	top[1] = 1;
	dfs(1);
	for(int i = 1; i <= n; i++) w[i] += w[i - 1];
	work(1);
	while(q--)
	{
		ll vi = read() ^ ans, vj = read() ^ ans, k = read() ^ ans;
		int l = LCA(vi, vj);
		ans = Query(vi, k) + Query(vj, k) - Query(l, k) - Query(fa[l], k);
		cout << ans << "\n";
		ans *= type;
	}
	return 0;
}