Count On A Tree II.

$n$ 个点的树，数一条链上有多少不同的点

sol：

树上莫队

首先，王室联邦分块

记 $(cu,cv)$ 为当前的链，$(qu,qv)$ 为当前询问的链，维护一个 $vis$ 数组表示“当前点在/不在当前链上”，每次暴力从 $cu,qu$ 爬到他们的 lca，从 $cv,qv$ 爬到他们的 lca，特盘一下 $qu,qv$ 的 lca 就可以了

#include <bits/stdc++.h>
#define LL long long
using namespace std;
#define rep(i, s, t) for (register int i = (s), i__end = (t); i <= i__end; ++i)
#define dwn(i, s, t) for (register int i = (s), i__end = (t); i >= i__end; --i)
inline int read() {
    int x = 0, f = 1; char ch = getchar();
    for (; !isdigit(ch); ch = getchar()) if (ch == '-') f = -f;
    for (; isdigit(ch); ch = getchar()) x = 10 * x + ch - '0';
    return x * f;
}
const int maxn = 540010;
int n, m, b[maxn], a[maxn], blk, bcnt;
vector<int> G[maxn];
int fa[maxn], dep[maxn];
namespace splca {
    int size[maxn], top[maxn];
    void dfs1(int x) {
        size[x] = 1;
        for(auto to : G[x]) {
            if(to == fa[x]) continue;
            fa[to] = x;
            dfs1(to); size[x] += size[to];
        }
    }
    void dfs2(int x, int col) {
        int k = 0; top[x] = col;
        for(auto to : G[x])
            if(to != fa[x] && size[to] > size[k]) k = to;
        if(!k) return;
        dfs2(k, col);
        for(auto to : G[x])
            if(to != fa[x] && to != k) dfs2(to, to);
    }
    int lca(int x, int y) {
        while(top[x] != top[y]) {
            if(dep[top[x]] < dep[top[y]]) swap(x, y);
            x = fa[top[x]]; 
        }
        return dep[x] < dep[y] ? x : y;
    }
}
void lca_init() {splca::dfs1(1); splca::dfs2(1, 1);}
int lca(int x, int y) {return splca::lca(x, y);}

int size[maxn], bl[maxn], q[maxn], top;
int dfs2(int x) {
    int cur = 0;
    for(auto to : G[x]) {
        if(to == fa[x]) continue;
        dep[to] = dep[x] + 1;
        cur += dfs2(to);
        if(cur >= blk) {
            while(cur--) bl[q[--top]] = bcnt;
            bcnt++;
        }
    }
    q[++top] = x;
    return cur + 1;
}
int ans[maxn], vis[maxn], inq[maxn];
struct Ques {
    int u, v, fl, fr, id;
    bool operator < (const Ques &b) const {
        return fl == b.fl ? fr < b.fr : fl < b.fl;
    }
}qs[maxn];
int now;
void move(int &x) {
    if(inq[x]){
        if(--vis[a[x]] == 0) now--;
    }
    else if(++vis[a[x]] == 1) now++;
    inq[x] ^= 1;
    x = fa[x];
}
int main() {
    n = read(), m = read();
    blk = sqrt(n);
    rep(i, 1, n) b[i] = a[i] = read();
    sort(b + 1, b + n + 1);
    rep(i, 1, n) a[i] = lower_bound(b+1, b+n+1, a[i]) - b;
    rep(i, 2, n) {
        int u = read(), v = read();
        G[u].push_back(v); G[v].push_back(u);
    } lca_init(); dep[1] = 1; dfs2(1);
    while(top) bl[q[--top]] = bcnt;
    rep(i, 1, m) {
        int v = read(), u = read();
        if(bl[v] > bl[u]) swap(u, v);
        qs[i] = (Ques){v, u, bl[v], bl[u], i};
        //cout << v << " " << u << " " << bl[v] << " " << bl[u] << endl;
    }
    sort(qs + 1, qs + m + 1);
    int cu = 1, cv = 1;
    rep(i, 1, m) {
        int nu = qs[i].u, nv = qs[i].v;
        int anc = lca(cu, nu);
        while(cu != anc) move(cu);
        while(nu != anc) move(nu);
        anc = lca(cv, nv);
        while(cv != anc) move(cv);
        while(nv != anc) move(nv);
        cv = qs[i].v, cu = qs[i].u;
        anc = lca(cv, cu);
        ans[qs[i].id] = now + (!vis[a[anc]]);
    }
    rep(i, 1, m) printf("%d\n",ans[i]);
}