看到红蓝两维偏序，马上醒过来了。
然后发现还有一维偏序是子树，这个可以通过拍扁树得到 dfn 序得到。
然后子树转化为区间，可以红蓝两维排序、归并解决掉，而第三维 dfn 通过树状数组维护区间和。

#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 15, M = N << 1;
int n;
int h[N], e[M], ne[M], idx = 0;
void add(int a, int b) {
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

struct node {
    int id, dfn, sz, red, blue;
    //blue 写成 bule 让我想起了一位故人。
} a[N], b[N];
bool cmp(node a, node b) {
    if (a.red != b.red) return a.red < b.red;
    if (a.blue != b.blue) return a.blue < b.blue;
    return a.dfn > b.dfn;
}

int dfn[N], sz[N], tot = 0;

void dfs(int u, int father) {
    sz[u] = 1, dfn[u] = ++tot;
    for (int i = h[u]; ~i; i = ne[i]) {
        int v = e[i];
        if (v == father) continue;
        dfs(v, u);
        sz[u] += sz[v];
    }
}

int tr[N];
void change(int x, int y) {
    for ( ; x < N; x += x & -x) tr[x] += y;
}
int query(int x) {
    int res = 0;
    for ( ; x ; x -= x & -x) res += tr[x];
    return res;
}

int ans[N];
void cdq(int l, int r) {
    if (l == r) return;
    int mid = l + r >> 1;
    cdq(l, mid), cdq(mid + 1, r);
    for (int i = l, j = mid + 1, k = l; k <= r; k++) {
        if ((j > r) || (i <= mid && a[i].blue <= a[j].blue)) change(a[i].dfn, 1), b[k] = a[i++];
        else {
            ans[a[j].id] += query(a[j].dfn + a[j].sz - 1) - query(a[j].dfn - 1);
            b[k] = a[j++];
        }
    }
    for (int i = l; i <= mid; i++) change(a[i].dfn, -1);
    for (int i = l; i <= r; i++) a[i] = b[i];
}

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) h[i] = -1;
    for (int i = 1, u, v; i < n; i++) {
        scanf("%d%d", &u, &v);
        add(u, v), add(v, u);
    }
    dfs(1, 0);
    for (int i = 1, x, y; i <= n; i++) scanf("%d%d", &x, &y), a[i] = (node){i, dfn[i], sz[i], x, y};
    sort(a + 1, a + 1 + n, cmp);
    cdq(1, n);
    for (int i = 1; i <= n; i++)
        if (ans[i]) printf("%d\n", ans[i]);
    return 0;
}