小 Q 与树

给定一棵带权的树，每条边的距离都为 1 1 1，要我们求 ∑ u = 1 n ∑ v = 1 n m i n ( a u , a v ) d i s ( u , v ) \sum\limits_{u = 1} ^{n} \sum\limits_{v = 1} ^{n}min(a_u, a_v)dis(u, v) u=1∑n​v=1∑n​min(au​,av​)dis(u,v)，
m i n ( a u , a v ) d i s ( u , v ) = m i n ( a u , a v ) ( d e p [ u ] + d e p [ v ] − 2 × d e p [ l c a ( u , v ) ] ) min(a_u, a_v) dis(u, v) = min(a_u, a_v)\left(dep[u] + dep[v] - 2 \times dep[lca(u, v)]\right)\\ min(au​,av​)dis(u,v)=min(au​,av​)(dep[u]+dep[v]−2×dep[lca(u,v)])

如果考虑 dsu on tree，则是枚举 u u u，分两种情况统计答案：

• a u ≤ a v a_u \leq a_v au​≤av​，则 ∑ v ∈ S a u ( d e p [ u ] + d e p [ v ] − 2 × d e p [ l c a ( u , v ) ] ) \sum\limits_{v \in S} a_u(dep[u] + dep[v] - 2 \times dep[lca(u, v)]) v∈S∑​au​(dep[u]+dep[v]−2×dep[lca(u,v)])，则我们只要知道集合 S S S中有多少个点，以及 ∑ v ∈ S d e p [ v ] \sum\limits_{v \in S} dep[v] v∈S∑​dep[v]即可，

  设点的个数为 t o t a l total total， ∑ v ∈ S d e p [ v ] = S u m d e p \sum\limits_{v \in S} dep[v] = Sum_{dep} v∈S∑​dep[v]=Sumdep​，则上式等价于 a u × t a t a l × ( d e p [ u ] − 2 × d e p [ l c a ( u , v ) ] ) + a u × S u m d e p a_u \times tatal \times (dep[u] - 2 \times dep[lca(u, v)]) + a_u \times Sum_{dep} au​×tatal×(dep[u]−2×dep[lca(u,v)])+au​×Sumdep​。

• a u > a v a_u > a_v au​>av​，则 ∑ u ∈ S a v ( d e p [ u ] + d e p [ v ] − 2 × d e p [ l c a ( u , v ) ] ) \sum\limits_{u \in S} a_v(dep[u] + dep[v] - 2 \times dep[lca(u, v)]) u∈S∑​av​(dep[u]+dep[v]−2×dep[lca(u,v)])，则我们只要知道 S u m a v × d e p [ v ] Sum_{a_v \times dep[v]} Sumav​×dep[v]​，以及 S u m a v Sum_{a_v} Sumav​​即可求得答案，

  上式等价于 S u m a v × d e p [ v ] + S u m a v × ( d e p [ u ] − 2 × d e p [ l c a ( u , v ) ] ) Sum_{a_v \times dep[v]} + Sum_{a_v} \times (dep[u] - 2 \times dep[lca(u, v)]) Sumav​×dep[v]​+Sumav​​×(dep[u]−2×dep[lca(u,v)])。

所以可以对点权离散化，然后用线段树来维护上面需要的四个值，即可进行 dsu on tree，整体复杂度 n log ⁡ n log ⁡ n n \log n \log n nlognlogn。

由于上面的统计我们都是进行的单向计算，所以还要对上述计算完后的答案乘以 2 2 2即可。

#include <bits/stdc++.h>
#define ls rt << 1
#define rs rt << 1 | 1
#define mid (l + r >> 1)
#define lson ls, l, mid
#define rson rs, mid + 1, r

using namespace std;

const int N = 2e5 + 10, mod = 998244353;

int head[N], to[N << 1], nex[N << 1], cnt = 1;

int son[N], sz[N], l[N], r[N], rk[N], dep[N], tot;

int sum1[N << 2], sum2[N << 2], sum3[N << 2], sum4[N << 2];

int a[N], b[N], n, m;

inline int add(int x, int y) {
  return x + y < mod ? x + y : x + y - mod;
}

inline int sub(int x, int y) {
  return x >= y ? x - y : x - y + mod;
}

inline int mul(int x, int y) {
  return 1ll * x * y % mod;
}

void Add(int x, int y) {
  to[cnt] = y;
  nex[cnt] = head[x];
  head[x] = cnt++;
}

void dfs(int rt, int fa) {
  dep[rt] = dep[fa] + 1, sz[rt] = 1, l[rt] = ++tot, rk[tot] = rt;
  for (int i = head[rt]; i; i = nex[i]) {
    if (to[i] == fa) {
      continue;
    }
    dfs(to[i], rt);
    sz[rt] += sz[to[i]];
    if (!son[rt] || sz[to[i]] > sz[son[rt]]) {
      son[rt] = to[i];
    }
  }
  r[rt] = tot;
}

void push_up(int rt) {
  sum1[rt] = add(sum1[ls], sum1[rs]);
  sum2[rt] = add(sum2[ls], sum2[rs]);
  sum3[rt] = add(sum3[ls], sum3[rs]);
  sum4[rt] = add(sum4[ls], sum4[rs]);
}

void update(int rt, int l, int r, int x, int v, int op) {
  if (l == r) {
    if (op == 1) {
      sum1[rt] += 1, sum2[rt] = add(sum2[rt], v), sum3[rt] = add(sum3[rt], mul(b[x], v)), sum4[rt] = add(sum4[rt], b[x]);
    }
    else {
      sum1[rt] -= 1, sum2[rt] = sub(sum2[rt], v), sum3[rt] = sub(sum3[rt], mul(b[x], v)), sum4[rt] = sub(sum4[rt], b[x]);
    }
    return ;
  }
  if (x <= mid) {
    update(lson, x, v, op);
  }
  else {
    update(rson, x, v, op);
  }
  push_up(rt);
}

int ans, ans1, ans2, ans3, ans4, ans5;

void query(int rt, int l, int r, int L, int R) {
  if (l >= L && r <= R) {
    ans1 = add(ans1, sum1[rt]), ans2 = add(ans2, sum2[rt]), ans3 = add(ans3, sum3[rt]), ans4 = add(ans4, sum4[rt]);
    return ;
  }
  if (L <= mid) {
    query(lson, L, R);
  }
  if (R > mid) {
    query(rson, L, R);
  }
}

void dfs(int rt, int fa, bool keep) {
  for (int i = head[rt]; i; i = nex[i]) {
    if (to[i] == fa || to[i] == son[rt]) {
      continue;
    }
    dfs(to[i], rt, 0);
  }
  if (son[rt]) {
    dfs(son[rt], rt, 1);
  }
  for (int i = head[rt]; i; i = nex[i]) {
    if (to[i] == fa || to[i] == son[rt]) {
      continue;
    }
    for (int j = l[to[i]]; j <= r[to[i]]; j++) {
      ans1 = ans2 = ans3 = ans4 = 0;
      query(1, 1, m, a[rk[j]], m);
      ans = add(ans, mul(ans2, b[a[rk[j]]]));
      ans = add(ans, mul(b[a[rk[j]]], mul(ans1, sub(dep[rk[j]], 2 * dep[rt]))));
      if (a[rk[j]] != 1) {
        ans1 = ans2 = ans3 = ans4 = 0;
        query(1, 1, m, 1, a[rk[j]] - 1);
        ans = add(ans, ans3);
        ans = add(ans, mul(ans4, sub(dep[rk[j]], 2 * dep[rt])));
      }
    }
    for (int j = l[to[i]]; j <= r[to[i]]; j++) {
      update(1, 1, m, a[rk[j]], dep[rk[j]], 1);
    }
  }
  ans1 = ans2 = ans3 = ans4 = 0;
  query(1, 1, m, a[rt], m);
  ans = add(ans, mul(ans2, b[a[rt]]));
  ans = add(ans, mul(b[a[rt]], mul(ans1, sub(dep[rt], 2 * dep[rt]))));
  if (a[rt] != 1) {
    ans1 = ans2 = ans3 = ans4 = 0;
    query(1, 1, m, 1, a[rt] - 1);
    ans = add(ans, ans3);
    ans = add(ans, mul(ans4, sub(dep[rt], 2 * dep[rt])));
  }
  update(1, 1, m, a[rt], dep[rt], 1);
  if (!keep) {
    for (int i = l[rt]; i <= r[rt]; i++) {
      update(1, 1, m, a[rk[i]], dep[rk[i]], -1);
    }
  }
}

int main() {
  // freopen("in.txt", "r", stdin);
  // freopen("out.txt", "w", stdout);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
  }
  sort(b + 1, b + 1 + n);
  m = unique(b + 1, b + 1 + n) - (b + 1);
  for (int i = 1; i <= n; i++) {
    a[i] = lower_bound(b + 1, b + 1 + m, a[i]) - b;
  }
  for (int i = 1, x, y; i < n; i++) {
    scanf("%d %d", &x, &y);
    Add(x, y);
    Add(y, x);
  }
  dfs(1, 0);
  dfs(1, 0, 1);
  printf("%d\n", mul(2, ans));
  return 0;
}