小 Q 与树（dsu on tree + segment tree）牛客练习赛 81 D

小 Q 与树

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

如果考虑 dsu on tree，则是枚举 $u$ ，分两种情况统计答案：

$au≤ava_u \leq a_v$ ，则 $∑v∈Sau(dep[u]+dep[v]−2×dep[lca(u,v)])\sum\limits_{v \in S} a_u(dep[u] + dep[v] - 2 \times dep[lca(u, v)])$ ，则我们只要知道集合 $S$ 中有多少个点，以及 $∑v∈Sdep[v]\sum\limits_{v \in S} dep[v]$ 即可，

设点的个数为 $t o t a l$ ， $∑v∈Sdep[v]=Sumdep\sum\limits_{v \in S} dep[v] = Sum_{dep}$ ，则上式等价于 $au×tatal×(dep[u]−2×dep[lca(u,v)])+au×Sumdepa_u \times tatal \times (dep[u] - 2 \times dep[lca(u, v)]) + a_u \times Sum_{dep}$ 。
$a_u > a_v$ ，则 $∑u∈Sav(dep[u]+dep[v]−2×dep[lca(u,v)])\sum\limits_{u \in S} a_v(dep[u] + dep[v] - 2 \times dep[lca(u, v)])$ ，则我们只要知道 $Sumav×dep[v]Sum_{a_v \times dep[v]}$ ，以及 $Sum_{a_v}$ 即可求得答案，

上式等价于 $Sumav×dep[v]+Sumav×(dep[u]−2×dep[lca(u,v)])Sum_{a_v \times dep[v]} + Sum_{a_v} \times (dep[u] - 2 \times dep[lca(u, v)])$ 。

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 $\log n \log n$ 。

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 $2$ 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, rusing 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;
}