Bzoj 4551: [Tjoi2016&Heoi2016]树

题面

戳我

Sol

每个被标记的点只会影响到它的子树，那么直接用线段树在dfn上搞
单点查询，区间修改

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1e5 + 10);

IL ll Read(){
    RG ll x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int n, Q, fst[_], nxt[_], to[_], cnt, fa[_], dfn[_], deep[_], Index, id[_], ed[_];
int mx[_ << 2];

IL void Add(RG int u, RG int v){  to[cnt] = v; nxt[cnt] = fst[u]; fst[u] = cnt++;  }

IL void Dfs(RG int u, RG int ff){
    fa[u] = ff; deep[u] = deep[ff] + 1; dfn[u] = ++Index;
    for(RG int e = fst[u]; e != -1; e = nxt[e]) Dfs(to[e], u);
	ed[u] = Index;
}

IL void Update(RG int x, RG int y){	 if(deep[mx[y]] > deep[mx[x]]) mx[x] = mx[y];  }

IL void Modify(RG int x, RG int l, RG int r, RG int L, RG int R, RG int p){
	if(L <= l && R >= r){  if(deep[p] > deep[mx[x]]) mx[x] = p;	return;  }
	RG int mid = (l + r) >> 1;
	if(mx[x]) Update(x << 1, x), Update(x << 1 | 1, x), mx[x] = 0;
	if(L <= mid) Modify(x << 1, l, mid, L, R, p);
	if(R > mid) Modify(x << 1 | 1, mid + 1, r, L, R, p);
}

IL int Query(RG int x, RG int l, RG int r, RG int p){
	if(l == r) return mx[x];
	RG int mid = (l + r) >> 1;
	if(mx[x]) Update(x << 1, x), Update(x << 1 | 1, x), mx[x] = 0;
	if(p <= mid) return Query(x << 1, l, mid, p);
	return Query(x << 1 | 1, mid + 1, r, p);
}

int main(RG int argc, RG char* argv[]){
    n = Read(); Q = Read(); Fill(fst, -1);
	for(RG int i = 1, x, y; i < n; ++i) x = Read(), y = Read(), Add(x, y);
	Dfs(1, 0); mx[1] = 1;
	for(RG int x; Q; --Q){
		RG char op; scanf(" %c", &op); x = Read();
		if(op == 'C')  Modify(1, 1, n, dfn[x], ed[x], x);
		else printf("%d\n", Query(1, 1, n, dfn[x]));
	}
    return 0;
}