樹上莫隊模板

題目鏈接：https://vjudge.net/problem/SPOJ-COT2

教學博客：https://www.cnblogs.com/zwfymqz/p/9223425.html

題目大意：查詢鏈上點權不同數的個數。

ps：如果是邊權的話，那麼u,v對應的區間是[st[u]+1,st[v]]，無需lca。

#include <bits/stdc++.h>
#define rep(i, a, b) for(int i = (a); i <= (b); i++)
#define pb push_back
#define all(x) (x).begin(),(x).end()
using namespace std;
const int N = 4e4+1;        //點數
const int M = 1e5+1;
int st[N],cnt,euler[N*2],deep[N],ed[N],fa[N][20];
vector<int> nxt[N];
void dfs(int u,int f) {
    deep[u] = deep[f]+1;
    fa[u][0] = f;
    euler[++cnt] = u;
    st[u] = cnt;
    for(auto v:nxt[u]) {
        if(v==f) continue;
        dfs(v,u);
    }
    euler[++cnt] = u;
    ed[u] = cnt;
}
void init(int n) {
    for(int j = 1; j <= 19; j++)
        for(int i = 1; i <= n; i++)
            fa[i][j] = fa[fa[i][j-1]][j-1];
}
int lca(int u, int v){
    if(deep[u] < deep[v]) swap(u, v);
    int diff = deep[u] - deep[v];
    for(int i = 19; i >= 0; i--) if(diff>>i&1) u = fa[u][i];
    if(u == v) return u;
    for(int i = 19; i >= 0; i--) if(fa[u][i] != fa[v][i]) u = fa[u][i],v = fa[v][i];
    return fa[u][0];
}
//-------------------------------------------------------
int ans[M],block,Ans,num[M],s[N],bid[2*N],bnum;
bool vis[M];
struct node{
    int l,r,lca,id;
    bool operator < (const node &cmp) const{ //常數優化
         return (bid[l]^bid[cmp.l])?(bid[l]<bid[cmp.l]):((bid[l]&1)?r<cmp.r:r>cmp.r);
    }
}q[M];
void add(int x){
    if(++num[x]==1) Ans++;
}
void del(int x){
    if(--num[x]==0) Ans--;
}
void Add(int p) {
    if(vis[p]) del(s[p]);  //判斷節點p出現奇偶次，奇數在鏈上，偶數不在鏈上
    else add(s[p]);
    vis[p] ^= 1;
}
//-------------------------------------------------------
int n,m,u,v;
vector<int>k;
int main() {	
    ios::sync_with_stdio(0);
    cin>>n>>m;
    rep(i, 1, n) cin>>s[i],k.pb(s[i]);
    sort(all(k));           //點權離散化
    k.erase(unique(all(k)),k.end());
    rep(i, 1, n) s[i] = lower_bound(all(k),s[i])-k.begin()+1;

    rep(i, 1, n-1) {
        cin>>u>>v;
        nxt[u].pb(v);
        nxt[v].pb(u);
    }
    dfs(1,0);
    init(n);

    block = sqrt(2*n);
    bnum = ceil(double(2*n)/block);
    rep(i, 1, bnum) rep(j, block*(i-1)+1, min(2*n,i*block)) bid[j] = i;  //常數優化

    rep(i, 1, m) {
        cin>>u>>v;
        if(st[u]>st[v]) swap(u,v);
        int _lca = lca(u,v);
        if(_lca==u) q[i] = (node){st[u],st[v],0,i};
        else q[i] = (node){ed[u],st[v],_lca,i};
    }

    sort(q+1,q+m+1);
    int l = 1,r = 0;
    rep(i, 1, m){
        while(r < q[i].r) Add(euler[++r]);
        while(r > q[i].r) Add(euler[r--]);
        while(l < q[i].l) Add(euler[l++]);
        while(l > q[i].l) Add(euler[--l]);
        if(q[i].lca) Add(q[i].lca);		//特判lca
        ans[q[i].id] = Ans;
        if(q[i].lca) Add(q[i].lca);
    }
    rep(i, 1, m) printf("%d\n",ans[i]);
    return 0;
}