「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;
}