Submission #789546

#	Time	Username	Problem	Language	Result	Execution time	Memory
789546		MISM06	Sumtree (INOI20_sumtree)	C++14	100 / 100	420 ms	111532 KiB

Main

//0 1 1 0 1
//0 1 0 0 1
//1 0 0 1 1
//0 1 1 0 1
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops")
// #pragma GCC target("avx2")

using namespace std;

#define F 			first
#define S 			second
#define pb 			push_back
#define sze			size()
#define	all(x)		x.begin() , x.end()
#define wall__		cout << "--------------------------------------\n";
#define kids		int mid = (tl + tr) >> 1, cl = v << 1, cr = v << 1 | 1
#define file_io		freopen("input.cpp", "r", stdin); freopen("output.cpp", "w", stdout);

typedef long long ll;
typedef long double dl;
typedef pair < int , int > pii;
typedef pair < int , ll > pil;
typedef pair < ll , int > pli;
typedef pair < ll , ll > pll;
typedef pair < int , pii > piii;
typedef pair < ll, pll > plll;


const ll N = 5e5 + 10, M = 6e5 + 10;
const ll mod = 1e9 + 7;
const ll inf = 2e16;
const ll INF = 1e9 + 10;
const ll lg = 32;

ll fac[M], inv[M];
inline void add (ll &x, ll y) {
	x += y; if (x >= mod) x -= mod;
	if (x < 0) x += mod;
}
inline ll Sum (ll x, ll y) {
	x += y; if (x >= mod) x -= mod;
	if (x < 0) x += mod;
	return x;
}
inline ll mul (ll x, ll y) {
	x *= y; x %= mod; if (x < 0) x+= mod;
	return x;
}
inline ll pw (ll x, ll y) {
	ll res = 1;
	while(y) {
		if (y & 1) res = mul(res, x);
		x = mul(x, x);
		y >>= 1;
	}
	return res;
}
inline ll choose (ll x, ll y) {
	if (x < y || y < 0) return 0;
	return mul(fac[x], mul(inv[y], inv[x - y]));
}

ll n, chp[N], par[N], hv[N], h[N], sub[N], rst[N];
ll a[N], st[N], en[N], timer = 0, ct[N], c0;
set < int > s[N];
vector < int > g[N];

struct segtree {
	ll seg[N << 2];
	void build (int v = 1, int tl = 1, int tr = n) {
		if (tl == tr) {
			seg[v] = 1;
			return;
		} kids;
		build(cl, tl, mid);
		build(cr, mid + 1, tr);
		seg[v] = mul(seg[cl], seg[cr]);
	}
	void upd (int idx, ll val, int v = 1, int tl = 1, int tr = n) {
		if (tl == tr) {
			seg[v] = val;
			return;
		} kids;
		if (idx <= mid) upd(idx, val, cl, tl, mid);
		else upd(idx, val, cr, mid + 1, tr);
		seg[v] = mul(seg[cl], seg[cr]);
	}
	ll ask () {
		return seg[1];
	}
} ans;

struct fen {
	ll bit[N];
	fen () {
		fill(bit, bit + N, 0);
	}
	inline ll read (int idx) {
		ll res = 0;
		for (; idx > 0; idx -= (idx & -idx)) res += bit[idx];
		return res;
	}
	inline void upd (int idx, ll val) {
		for (; idx <= n; idx += (idx & -idx)) 
			bit[idx] += val;
	}
	inline ll range (int l, int r) {
		return read(r) - read(l - 1);
	}
} f1, f2;
void dfs0 (int v, int p) {
	par[v] = p;
	h[v] = h[p] + 1;
	sub[v] = 1;
	for (auto u : g[v]) {
		if (u == p) continue;
		dfs0(u, v);
		sub[v] += sub[u];
		if (sub[u] > sub[hv[v]]) hv[v] = u;
	}
}
void dfs (int v, int p) {
	st[v] = ++timer; rst[st[v]] = v;
	if (hv[v] != 0) {
		chp[hv[v]] = chp[v];
		dfs(hv[v], v);
	}
	for (auto u : g[v]) {
		if (u == p || hv[v] == u) continue;
		chp[u] = u;
		dfs(u, v);
	}
	en[v] = timer;
}
void out () {
	cout << ans.ask() << '\n';
}
bool is[N];
int find_par (int v) {
	int p = chp[v];
	if (s[p].sze != 0) {
		auto it = s[p].lower_bound(st[v]);
		if (it != s[p].begin()) {
			--it;
			return rst[*it];
		}
	}
	v = par[p];
	p = chp[par[p]];
	while (true) {

		if (s[p].sze != 0) {
			auto it = s[p].upper_bound(st[v]);
			if (it != s[p].begin()) {
				--it;
				return rst[*it];
			}
		}
		v = par[p];
		p = chp[par[p]];
	}
	// v = par[v];
	// while(!is[v]) {
	// 	v = par[v];
	// }
	// return v;	
}

void solve () {

	cin >> n >> a[1];
	for (int i = 1; i < n; i++) {
		int v, u; cin >> v >> u;
		g[v].pb(u); g[u].pb(v);
	}
	sub[0] = -1;
	ans.build();
	chp[1] = 1;
	dfs0(1, 0);
	dfs(1, 0);
	ct[1] = n; f1.upd(st[1], n); f2.upd(st[1], a[1]);
	ll rs = choose(a[1] + ct[1] - 1, ct[1] - 1);
	ans.upd(1, rs);
	s[chp[1]].insert(st[1]);
	is[1] = 1;
	out();
	int q; cin >> q;
	while (q--) {
		int t; cin >> t;
		if (t == 1) {
			ll v, x; cin >> v >> x;
			ct[v] = sub[v] - f1.range(st[v], en[v]);
			f1.upd(st[v], ct[v]);
			a[v] = x - f2.range(st[v], en[v]);
			f2.upd(st[v], a[v]);
			rs = choose(a[v] + ct[v] - 1, ct[v] - 1);
			ans.upd(v, rs);
			s[chp[v]].insert(st[v]);
			is[v] = 1;

			int p = find_par(v);
			// cout << "p = " << p << '\n';
			ct[p] -= ct[v];
			f1.upd(st[p], -ct[v]);
			a[p] -= a[v];
			f2.upd(st[p], -a[v]);
			rs = choose(a[p] + ct[p] - 1, ct[p] - 1);
			ans.upd(p, rs);
			out();

		} else {
			int v; cin >> v;
			f1.upd(st[v], -ct[v]);
			f2.upd(st[v], -a[v]);
			rs = 1;
			ans.upd(v, rs);
			s[chp[v]].erase(st[v]);
			is[v] = 0;

			int p = find_par(v);
			// cout << "p = " << p << '\n';
			ct[p] += ct[v];
			f1.upd(st[p], ct[v]);
			a[p] += a[v];
			f2.upd(st[p], a[v]);
			rs = choose(a[p] + ct[p] - 1, ct[p] - 1);
			ans.upd(p, rs);
			out();
		}
	}


}


int main() {
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	fac[0] = 1;
	for (int i = 1; i < M; i++) fac[i] = mul(fac[i - 1], i);
	inv[M - 1] = pw(fac[M - 1], mod - 2);
	for (int i = M - 2; i >= 0; --i) inv[i] = mul(inv[i + 1], i + 1);
	int t = 1;
	// cin >> t;
	while (t--) {solve();}
    return 0;
}
/*
10 10
1 2
2 4
2 5
5 7
5 8
1 3
3 6
6 9
6 10
13
1 2 10
1 5 10
1 7 10
1 8 0
2 2
1 2 0
2 7
2 2
1 7 0
2 7
1 7 3
2 8
2 5


92378
1001
66
1
1
1
0
0
11
1
11
1
8
6435
*/
//shrek is love;

• Subtask 1: 10 / 10
• Subtask 2: 20 / 20
• Subtask 3: 20 / 20
• Subtask 4: 20 / 20
• Subtask 5: 30 / 30