loj#2049. 「HNOI2016」网络(set 树剖 暴力)

题意

题目链接

Sol

下面的代码是\(O(nlog^3n)\)的暴力。

因为从一个点向上只会跳\(logn\)次，所以可以暴力的把未经过的处理出来然后每个点开个multiset维护最大值

#include<bits/stdc++.h> 
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long 
#define LL long long 
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 2e5 + 10, SS = MAXN * 4, mod = 1e9 + 7, INF = 1e9 + 10;
const double eps = 1e-9;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename A> inline void debug(A a){cout << a << '\n';}
template <typename A> inline LL sqr(A x){return 1ll * x * x;}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, Q, fa[MAXN], siz[MAXN], son[MAXN], id[MAXN], top[MAXN], dep[MAXN], times;
vector<int> v[MAXN];
void dfs1(int x, int _fa) {
	siz[x] = 1; dep[x] = dep[_fa] + 1; fa[x] = _fa;
	for(auto &to : v[x]) {
		if(to == _fa) continue;
		dfs1(to, x);
		siz[x] += siz[to];
		if(siz[to] > siz[son[x]]) son[x] = to;
	}
}
void dfs2(int x, int topf) {
	top[x] = topf; id[x] = ++times;
	if(!son[x]) return ;
	dfs2(son[x], topf);
	for(auto &to : v[x]) {
		if(top[to]) continue;
		dfs2(to, to);
	}
}
multiset<int> s[SS];
struct Query {
	int a, b, v;
}q[MAXN];
vector<Pair> line[MAXN];
int ls[SS], rs[SS], root, tot;
void Erase(multiset<int> &s, int v) {
	auto it = s.find(v);
	if(it != s.end()) s.erase(it);
}
void Get(vector<Pair> &v, int x, int y) {
	vector<Pair> tmp;
	while(top[x] ^ top[y]) {
		if(dep[top[x]] < dep[top[y]]) swap(x, y);
		tmp.push_back({id[top[x]], id[x]});
		x = fa[top[x]];
	}
	if(dep[x] > dep[y]) swap(x, y);
	tmp.push_back({id[x], id[y]});
	sort(tmp.begin(), tmp.end());
	int las = 1;
	for(auto x : tmp) {
		if(las <= x.fi - 1) v.push_back({las, x.fi - 1});
		las = x.se + 1;
	}
	if(las <= N) v.push_back({las, N});
}

int Mx(multiset<int> &s) {
	if(s.empty()) return -1;
	auto it = s.end(); it--;
	return *it;
}
void IntAdd(int &k, int l, int r, int ql, int qr, int v, int opt) {
	if(!k) k = ++tot;
	if(ql <= l && r <= qr) {
		if(opt == 1) s[k].insert(v); 
		else Erase(s[k], v);
		return ;
	}
	int mid = l + r >> 1;
	if(ql <= mid) IntAdd(ls[k], l, mid, ql, qr, v, opt);
	if(qr  > mid) IntAdd(rs[k], mid + 1, r, ql, qr, v, opt);
}
int Query(int k, int l, int r, int p) {
	if(!k) return -1;
	int ans = Mx(s[k]), mid = l + r >> 1;
	if(l == r) return Mx(s[k]);
	if(p <= mid) chmax(ans, Query(ls[k], l, mid, p));
	else chmax(ans, Query(rs[k], mid + 1, r, p));
	return ans;
}
void TreeAdd(int x, int y, int v, int opt) {
	while(top[x] ^ top[y]) {
		if(dep[top[x]] < dep[top[y]]) swap(x, y);
		IntAdd(root, 1, N, id[top[x]], id[x], v, opt);
		x = fa[top[x]];
	}
	if(dep[x] > dep[y]) swap(x, y);
	IntAdd(root, 1, N, id[x], id[y], v, opt);
}
void Add(int ti, int opt) {
	int x = q[ti].a, y = q[ti].b, v = q[ti].v;
	if(opt == 1) Get(line[ti], x, y);
	for(auto x : line[ti]) 
		IntAdd(root, 1, N, x.fi, x.se, v, opt);
}
signed main() {
//	Fin(a); Fout(b);
    N = read(); Q = read();
    for(int i = 1; i <= N - 1; i++) {
    	int x = read(), y = read();
    	v[x].push_back(y);
    	v[y].push_back(x);
	}
    dfs1(1, 0);
    dfs2(1, 1);
	for(int i = 1; i <= Q; i++) {
		int opt = read();
		if(opt == 0) {
			int a = read(), b = read(), v = read(); q[i] = {a, b, v};
			Add(i, 1); 
 		} else if(opt == 1) {
			int ti = read();
			Add(ti, -1); 
		} else if(opt == 2) {
			int x = read();
			printf("%d\n", Query(root, 1, N, id[x]));
		}
	}
	return 0;
}