meaning of the title

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Sol

Think of the chairman tree and heuristic merger. Obviously they can't.

Calculated as dfs in the order of violence. Bow to the wind

When it is implemented, the number of occurrences of (tim[i]) colors can be directly expressed by (tim[i] and (ans[i] by the number of occurrences of colors more than (i)

Since the left and right endpoints will only affect one(ans[i], the modification is(O(1)

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 10;
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;
}
int N, M, dfn[MAXN], rev[MAXN], tot, block, bel[MAXN], siz[MAXN], col[MAXN], tims[MAXN], Ans[MAXN], out[MAXN];
vector<int> v[MAXN];
struct Query{
    int id, l, r, k;
    bool operator < (const Query &rhs) const {
        return bel[l] == bel[rhs.l] ? r < rhs.r : bel[l] < bel[rhs.l];
    }
}Q[MAXN];
void dfs(int x, int fa) {
    dfn[x] = ++tot; rev[tot] = x; siz[x] = 1;
    for(int i = 0, to; i < v[x].size(); i++) {
        if((to = v[x][i]) == fa) continue;
        dfs(to, x); siz[x] += siz[to];
    }
}
void add(int x, int opt) {
    if(opt == 1) Ans[++tims[x]]++;
    else Ans[tims[x]--]--;
}
void solve() {  
    sort(Q + 1, Q + M + 1);
    int l = 1, r = 0;
    for(int i = 1; i <= M; i++) {
        while(r > Q[i].r) add(col[rev[r--]], -1);
        while(r < Q[i].r) add(col[rev[++r]], 1);
        while(l < Q[i].l) add(col[rev[l++]], -1);
        while(l > Q[i].l) add(col[rev[--l]], 1);
        out[Q[i].id] = Ans[Q[i].k];
        ///printf("%d\n", out[Q[i].id]);
    }
    for(int i = 1; i <= M; i++) printf("%d\n", out[i]);

}
int main() {
    N = read(); M = read(); block = sqrt(N);
    for(int i = 1; i <= N; i++) col[i] = read(), bel[i] = (i - 1) / block + 1;
    for(int i = 1; i <= N - 1; i++) {
        int x = read(), y = read();
        v[x].push_back(y); v[y].push_back(x);
    }   
    dfs(1, 0);
    for(int i = 1; i <= M; i++) {
        Q[i].id = i; int x = read(); Q[i].k = read();
        Q[i].l = dfn[x];
        Q[i].r = dfn[x] + siz[x] -1;
    }
    solve();
    return 0;
}
/*
*/