给定一颗以$1$号节点为根的树，如果$a\ne b$，且$a$是$b$的祖先，则$a$比$b$更厉害，如果$a\ne b$，且$dis(a,b)\le x$，$x$为给定的一个数，则$a,b$紧邻。

现有$m$次询问，每次询问给定$p,k$，为存在多少个三元组$(p,b,c)$满足一下条件：

• $p,b$都比$c$厉害。
• $p,b$彼此紧邻对于给定的常数$k$。

因为$p,b$都比$c$更厉害，则$p,b$都是$c$的祖先，分情况讨论：

• $p$是$b$的祖先

  只要满足$b$在$p$的子树中且，$c$在$b$的子树中即可。

• $b$是$p$的祖先

  $b$一定在$1->p$的路径上，且$c$在$p$的子树中。

线段树合并搞一搞，在线，离线都可，复杂度$O(n\log n)$。

#include <bits/stdc++.h>using namespace std;const int N = 3e5 + 10;int head[N], to[N << 1], nex[N << 1], cnt = 1;int root[N], ls[N << 5], rs[N << 5], num;int dep[N], sz[N], n, m;long long sum[N << 5];long long ans[N];vector<pair<int, int>> query[N];void add(int x, int y) {to[cnt] = y;nex[cnt] = head[x];head[x] = cnt++;
}int merge(int x, int y, int l, int r) {if (!x || !y) {return x | y;}if (l == r) {sum[x] += sum[y];return x;}int mid = l + r >> 1;ls[x] = merge(ls[x], ls[y], l, mid);rs[x] = merge(rs[x], rs[y], mid + 1, r);sum[x] = sum[ls[x]] + sum[rs[x]];return x;
}void update(int &rt, int l, int r, int x, int v) {if (!rt) {rt = ++num;}sum[rt] += v;if (l == r) {return ;}int mid = l + r >> 1;if (x <= mid) {update(ls[rt], l, mid, x, v);}else {update(rs[rt], mid + 1, r, x, v);}
}long long ask(int rt, int l, int r, int L, int R) {if (l >= L && r <= R) {return sum[rt];}long long ans = 0;int mid = l + r >> 1;if (L <= mid) {ans += ask(ls[rt], l, mid, L, R);}if (R > mid) {ans += ask(rs[rt], mid + 1, r, L, R);}return ans;
}void dfs(int rt, int fa) {dep[rt] = dep[fa] + 1, sz[rt] = 1;for (int i = head[rt]; i; i = nex[i]) {if (to[i] == fa) {continue;}dfs(to[i], rt);sz[rt] += sz[to[i]];root[rt] = merge(root[rt], root[to[i]], 1, n);}for (auto it : query[rt]) {int id = it.first, k = it.second;ans[id] = ask(root[rt], 1, n, dep[rt], min(dep[rt] + k, n));ans[id] += 1ll * (dep[rt] - max(1, dep[rt] - k)) * (sz[rt] - 1);}update(root[rt], 1, n, dep[rt], sz[rt] - 1);
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);scanf("%d %d", &n, &m);for (int i = 1, x, y; i < n; i++) {scanf("%d %d", &x, &y);add(x, y);add(y, x);}for (int i = 1, x, k; i <= m; i++) {scanf("%d %d", &x, &k);query[x].push_back({i, k});}dfs(1, 0);for (int i = 1; i <= m; i++) {printf("%lld\n", ans[i]);}return 0;
}