题意
Sol
首先对询问差分一下，我们就只需要统计(u, v, lca(u, v), fa[lca(u, v)])到根的路径的贡献。
再把每个点与(k)的lca的距离差分一下，则只需要统计每个点与(k)的lca深度。这个东西等价于所有的链与(k)到根的链的并。
树剖+主席树维护一下。这题的主席树需要区间加1，可以标记永久化合并标记
复杂度(O(nlog ^2n))
#include<bits/stdc++.h>
#define Pair pair<LL, LL>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long
#define LL long long
#define ull unsigned long long
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 2e5 + 10, mod = 1e9 + 7, INF = 1e9 + 10;
const double eps = 1e-9;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename A> inline void debug(A a){cout << a << '
';}
template <typename A> inline LL sqr(A x){return 1ll * x * x;}
template <typename A, typename B> inline LL fp(A a, B p, int md = mod) {int b = 1;while(p) {if(p & 1) b = mul(b, a);a = mul(a, a); p >>= 1;}return b;}
template <typename A> A inv(A x) {return fp(x, mod - 2);}
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
LL lastans;
int type, N, Q, p[MAXN], fa[MAXN], top[MAXN], dep[MAXN], son[MAXN], siz[MAXN], dfn[MAXN], rev[MAXN], times;
LL Esum[MAXN], sdis[MAXN], valE[MAXN];
vector<Pair> v[MAXN];
void dfs(int x, int _fa) {
fa[x] = _fa; dep[x] = dep[_fa] + 1; siz[x] = 1;
for(auto &tmp : v[x]) {
int to = tmp.first, w = tmp.se;
if(to == _fa) continue;
Esum[to] = Esum[x] + w; valE[to] = w;
dfs(to, x);
siz[x] += siz[to];
if(siz[to] > siz[son[x]]) son[x] = to;
}
}
void dfs2(int x, int topf) {
top[x] = topf; dfn[x] = ++times; rev[times] = x;
if(!son[x]) return ;
dfs2(son[x], topf);
for(auto &to : v[x]) {
if(top[to.fi]) continue;
dfs2(to.fi, to.fi);
}
}
int LCA(int x, int y) {
while(top[x] ^ top[y]) {
if(dep[top[x]] < dep[top[y]]) swap(x, y);
x = fa[top[x]];
}
return dep[x] < dep[y] ? x : y;
}
int rt[MAXN], ls[MAXN * 80], rs[MAXN * 80], cnt;
LL sumc[MAXN * 80], sum[MAXN * 80], lzy[MAXN * 80];
void Build(int &k, int l, int r) {
k = ++cnt;
if(l == r) {sumc[k] = valE[rev[l]]; return ;}
int mid = l + r >> 1;
Build(ls[k], l, mid); Build(rs[k], mid + 1, r);
sumc[k] = sumc[ls[k]] + sumc[rs[k]];
}
Pair operator + (const Pair a, const Pair b) {
return {a.fi + b.fi, a.se + b.se};
}
void Add(int &k, int l, int r, int ql, int qr) {
++cnt; int nw = cnt;
ls[nw] = ls[k]; rs[nw] = rs[k]; sumc[nw] = sumc[k];
sum[nw] = sum[k]; lzy[nw] = lzy[k];
k = cnt;
if(ql <= l && r <= qr) {
lzy[k]++, sum[k] += sumc[k];
return ;
}
int mid = l + r >> 1;
if(ql <= mid) Add(ls[k], l, mid, ql, qr);
if(qr > mid) Add(rs[k], mid + 1, r, ql, qr);
sum[k] = sum[ls[k]] + sum[rs[k]] + lzy[k] * sumc[k];
}
Pair Query(int k, int l, int r, int ql, int qr) {
if(ql <= l && r <= qr) return {sum[k], sumc[k]};
Pair res = {0, 0};
int mid = l + r >> 1;
if(ql <= mid) res = res + Query(ls[k], l, mid, ql, qr);
if(qr > mid) res = res + Query(rs[k], mid + 1, r, ql, qr);
res.fi += 1ll * res.se * lzy[k];
return res;
}
void insert(int x) {
int pre = x;
rt[x] = rt[fa[x]];
x = p[x];
while(x)Add(rt[pre], 1, N, dfn[top[x]], dfn[x]), x = fa[top[x]];
}
void dfs3(int x, int fa) {
sdis[x] = sdis[fa] + Esum[p[x]];
insert(x);
for(auto &to : v[x])
if(to.fi != fa) dfs3(to.fi, x);
}
LL solve(int u, int k) {
if(!k) return 0;
LL res = sdis[k] + 1ll * dep[k] * Esum[u];
while(u) {
res -= Query(rt[k], 1, N, dfn[top[u]], dfn[u]).fi << 1;
u = fa[top[u]];
}
return res;
}
signed main() {
type = read(); N = read(); Q = read();
for(int i = 1; i <= N - 1; i++) {
int x = read(),y = read(), w = read();
v[x].push_back({y, w});
v[y].push_back({x, w});
}
for(int i = 1; i <= N; i++) p[i] = read();
dfs(1, 0);
dfs2(1, 1);
Build(rt[0], 1, N);
dfs3(1, 0);
while(Q--) {
int tu = read() ^ (lastans * type), tv = read() ^ (lastans * type), tk = read() ^ (lastans * type);
lastans = solve(tk, tu) + solve(tk, tv) - solve(tk, LCA(tu, tv)) - solve(tk, fa[LCA(tu, tv)]);
cout << lastans << '
';
}
return 0;
}