【gym 102201F】:Fruit Tree 主席树 + LCA

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题意

给你一颗带权树，多次询问路径上出现次数过半的权值

分析

主席树维护一下$u$节点到根节点上，所有权值出现的次数即可
最后需要注意一下的是，在减去两倍$lca$的时候，会把$lca$一起减去，在查询过程过需要补上

代码

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 250010, M = N * 2;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {
	char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}
	while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
int h[N], e[M], ne[M], idx;
int father;
struct Node {
	int l, r;
	int cnt;
} tr[N * 100];
int root[N], dep[N], fa[N][25], a[N];
int n, m;

void add(int x, int y) {
	ne[idx] = h[x], e[idx] = y, h[x] = idx++;
}

int build(int l, int r) {
	int p = ++idx;
	tr[p].cnt = 0;
	if (l == r) return p;
	int mid = (l + r) >> 1;
	tr[p].l = build(l, mid), tr[p].r = build(mid + 1, r);
	return p;
}

int insert(int p, int l, int r, int x) {
	int q = ++idx;
	tr[q] = tr[p];
	if (l == r) {
		tr[q].cnt++;
		return q;
	}
	int mid = (l + r) >> 1;
	if (x <= mid) tr[q].l = insert(tr[p].l, l, mid, x);
	else tr[q].r = insert(tr[p].r, mid + 1, r, x);
	tr[q].cnt = tr[tr[q].l].cnt + tr[tr[q].r].cnt;
	return q;
}

void dfs(int u, int f) {
	fa[u][0] = f;
	dep[u] = dep[f] + 1;
	root[u] = insert(root[f], 1, n, a[u]);
	for (int i = 1; i <= 20; i++)
		fa[u][i] = fa[fa[u][i - 1]][i - 1];
	for (int i = h[u]; ~i; i = ne[i]) {
		int j = e[i];
		if (j == f) continue;
		dfs(j, u);
	}
}

int LCA(int a, int b) {
	if (dep[a] < dep[b]) swap(a, b);
	for (int i = 20; i >= 0; i--)
		if (dep[fa[a][i]] >= dep[b])
			a = fa[a][i];
	if (a == b) return a;
	for (int i = 20; i >= 0; i--)
		if (fa[a][i] != fa[b][i])
			a = fa[a][i], b = fa[b][i];
	return fa[a][0];
}

int query(int x,int y,int z,int l,int r,int k){
	int num = tr[y].cnt + tr[z].cnt - 2 * tr[x].cnt;
	if(l == r) return l;
	int cnt1 = tr[tr[y].l].cnt + tr[tr[z].l].cnt - 2 * tr[tr[x].l].cnt;
	int cnt2 = num - cnt1;
	
	int ans = -1;
	int mid = (l + r) >> 1;
	if(a[father] >= l && a[father] <= mid) cnt1++;
	else if(a[father] > mid && a[father] <= r) cnt2++;
	if(cnt1 >= k) ans = query(tr[x].l,tr[y].l,tr[z].l,l,mid,k);
	else if(cnt2 >= k) ans = query(tr[x].r,tr[y].r,tr[z].r,mid + 1,r,k);
	return ans; 
}

int main() {
	memset(h, -1, sizeof h);
	read(n), read(m);
	for (int i = 1; i <= n; i++) read(a[i]);
	for(int i = 1;i < n;i++){
		int a,b;
		read(a),read(b);
		add(a,b),add(b,a);
	}
	root[0] = build(1, n);
	dfs(1, 0);
	while(m--){
		int l,r;
		read(l),read(r);
		father = LCA(l,r);
		int num = dep[l] + dep[r] - 2 * dep[father] + 1;
		num = num / 2 + 1;
		int ans = query(root[father],root[l],root[r],1,n,num);
		di(ans);
	}
	return 0;
}