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Laijinyi

园龄：1年8个月粉丝：2关注：2

P3250 网络 题解

Solution

单次二分：问“重要度 ≥x 的所有操作，且 t 时间点还存在的所有操作中，是否有不经过这个点的”

整体二分：保持操作、询问按时间有序，即预先按时间排序，下传时保持有序；

对于一次 Solve，对于所有重要度 ≥mid+1 的操作（加入、删除），考虑与询问按时间混合排序，然后依次回答。

这里有比树剖更好的处理方法：树上差分，一次加入路径 (u,v) 就给 u,v 加一，lca,fa(lca) 减一；一次询问 u 就查 u 子树和是否等于操作数，树状数组可以实现。

用不用 O(1) LCA 都是 O(nlog2⁡n) 的.

Code 1

vector 版整体二分，未卡常

#include <bits/stdc++.h>
using namespace std;
#define rep(i, j, k) for (int i = (j); i <= (k); ++i)
#define reo(i, j, k) for (int i = (j); i >= (k); --i)
typedef long long ll;
const int N = 1e5 + 10, M = 2e5 + 10;
int n, m, tot, ans[N];
vector<int> G[N];
struct Item {
	int op, u, v, w, tim, id;
};
 
int tim, dfn[N], dep[N], sz[N], son[N], Top[N], Fa[N];
void DFS1(int u, int fa) {
	sz[u] = 1, Fa[u] = fa, dep[u] = dep[fa] + 1;
	for (int v : G[u])
		if (v != fa) {
			DFS1(v, u), sz[u] += sz[v];
			if (sz[v] > sz[son[u]]) son[u] = v;
		}
}
void DFS2(int u, int tp) {
	dfn[u] = ++tim, Top[u] = tp;
	if (son[u]) DFS2(son[u], tp);
	for (int v : G[u])
		if (v != Fa[u] && v != son[u])
			DFS2(v, v);
}
int LCA(int u, int v) {
	while (Top[u] != Top[v]) {
		if (dep[Top[u]] < dep[Top[v]]) swap(u, v);
		u = Fa[Top[u]];
	}
	return dep[u] < dep[v] ? u : v;
}
 
struct BIT {
	ll sum[N];
	BIT() {
		memset(sum, 0, sizeof(sum));
	}
	void Upd(int x, ll v) {
		for (; x <= n; x += x & -x) sum[x] += v;
	}
	ll Qry(int x) {
		ll res = 0;
		for (; x; x -= x & -x) res += sum[x];
		return res;
	}
	ll Qry(int x, int y) {
		return Qry(y) - Qry(x - 1);
	}
} bit;
 
void Solve(int l, int r, vector<Item> &Op) {
	int cntQ = 0;
	for (auto it : Op) cntQ += it.op == 2;
	if (!cntQ) return;
	if (l == r) {
		for (auto it : Op) {
			if (it.op == 2) ans[it.id] = (l == 1) ? -1 : (l - 1);
		}
		return;
	}
	int mid = (l + r) >> 1;
	vector<Item> OpL, OpR;
	int cnt = 0;
	for (auto it : Op) {
		if (it.op == 0) {
			if (it.w >= mid) {
				int u = it.u, v = it.v, lca = LCA(u, v), fa = Fa[lca];
				bit.Upd(dfn[u], 1), bit.Upd(dfn[v], 1), bit.Upd(dfn[lca], -1);
				if (fa) bit.Upd(dfn[fa], -1);
				++cnt;
			}
			if (it.w > mid) OpR.push_back(it);
			else OpL.push_back(it);
		}
		if (it.op == 1) {
			if (it.w >= mid) {
				int u = it.u, v = it.v, lca = LCA(u, v), fa = Fa[lca];
				bit.Upd(dfn[u], -1), bit.Upd(dfn[v], -1), bit.Upd(dfn[lca], 1);
				if (fa) bit.Upd(dfn[fa], 1);
				--cnt;
			}
			if (it.w > mid) OpR.push_back(it);
			else OpL.push_back(it);
		}
		if (it.op == 2) {
			int res = bit.Qry(dfn[it.u], dfn[it.u] + sz[it.u] - 1);
			if (res == cnt) {
				OpL.push_back(it);
			} else {
				OpR.push_back(it);
			}
		}
	}
	for (auto it : Op) {
		if (it.op == 0 && it.w >= mid) {
			int u = it.u, v = it.v, lca = LCA(u, v), fa = Fa[lca];
			bit.Upd(dfn[u], -1), bit.Upd(dfn[v], -1), bit.Upd(dfn[lca], 1);
			if (fa) bit.Upd(dfn[fa], 1);
		}
		if (it.op == 1 && it.w >= mid) {
			int u = it.u, v = it.v, lca = LCA(u, v), fa = Fa[lca];
			bit.Upd(dfn[u], 1), bit.Upd(dfn[v], 1), bit.Upd(dfn[lca], -1);
			if (fa) bit.Upd(dfn[fa], -1);
		}
	}
	Solve(l, mid, OpL), Solve(mid + 1, r, OpR);
}
 
int main() {
	ios::sync_with_stdio(false), cin.tie(nullptr);
	cin >> n >> m;
	rep(i, 1, n - 1) {
		int u, v;
		cin >> u >> v, G[u].push_back(v), G[v].push_back(u);
	}
	int mx = 0;
	vector<Item> Op(m);
	rep(i, 0, m - 1) {
		int x;
		cin >> Op[i].op;
		if (Op[i].op == 0) 
			cin >> Op[i].u >> Op[i].v >> Op[i].w, mx = max(mx, Op[i].w);
		if (Op[i].op == 1) 
			cin >> x, --x, Op[i].u = Op[x].u, Op[i].v = Op[x].v, Op[i].w = Op[x].w;
		if (Op[i].op == 2) 
			cin >> Op[i].u, Op[i].id = ++tot;
	}
	DFS1(1, 0), DFS2(1, 1);
	Solve(1, mx + 1, Op);
	rep(i, 1, tot) 
		cout << ans[i] << '\n';
	return 0;
}

Code 2

普通整体二分

#include <bits/stdc++.h>
using namespace std;
#define rep(i, j, k) for (int i = (j); i <= (k); ++i)
#define reo(i, j, k) for (int i = (j); i >= (k); --i)
typedef long long ll;
const int N = 1e5 + 10, M = 2e5 + 10;
int n, m, tot, ans[M];
vector<int> G[N];
struct Item {
	int op, u, v, w, tim, id;
} Op[M];
int id[M], _id[M], vis[M];
 
int tim, dfn[N], dep[N], sz[N], son[N], Top[N], Fa[N];
void DFS1(int u, int fa) {
	sz[u] = 1, Fa[u] = fa, dep[u] = dep[fa] + 1;
	for (int v : G[u])
		if (v != fa) {
			DFS1(v, u), sz[u] += sz[v];
			if (sz[v] > sz[son[u]]) son[u] = v;
		}
}
void DFS2(int u, int tp) {
	dfn[u] = ++tim, Top[u] = tp;
	if (son[u]) DFS2(son[u], tp);
	for (int v : G[u])
		if (v != Fa[u] && v != son[u])
			DFS2(v, v);
}
int LCA(int u, int v) {
	while (Top[u] != Top[v]) {
		if (dep[Top[u]] < dep[Top[v]]) swap(u, v);
		u = Fa[Top[u]];
	}
	return dep[u] < dep[v] ? u : v;
}
 
struct BIT {
	ll sum[N];
	BIT() {
		memset(sum, 0, sizeof(sum));
	}
	void Upd(int x, ll v) {
		for (; x <= n; x += x & -x) sum[x] += v;
	}
	ll Qry(int x) {
		ll res = 0;
		for (; x; x -= x & -x) res += sum[x];
		return res;
	}
	ll Qry(int x, int y) {
		return Qry(y) - Qry(x - 1);
	}
} bit;
 
void Solve(int l, int r, int ql, int qr) {
	int cntQ = 0;
	rep(i, ql, qr) cntQ += Op[id[i]].op == 2;
	if (!cntQ) return;
	if (l == r) {
		rep(i, ql, qr) {
			auto it = Op[id[i]];
			if (it.op == 2) ans[it.id] = (l == 1) ? -1 : (l - 1);
		}
		return;
	}
	int mid = (l + r) >> 1, cnt = 0, L = ql - 1, R;
	rep(i, ql, qr) {
		auto it = Op[id[i]];
		if (it.op == 0) {
			if (it.w >= mid) {
				int u = it.u, v = it.v, lca = LCA(u, v), fa = Fa[lca];
				bit.Upd(dfn[u], 1), bit.Upd(dfn[v], 1), bit.Upd(dfn[lca], -1);
				if (fa) bit.Upd(dfn[fa], -1);
				++cnt;
			}
			if (it.w > mid) vis[i] = 1;
			else _id[++L] = id[i], vis[i] = 0;
		}
		if (it.op == 1) {
			if (it.w >= mid) {
				int u = it.u, v = it.v, lca = LCA(u, v), fa = Fa[lca];
				bit.Upd(dfn[u], -1), bit.Upd(dfn[v], -1), bit.Upd(dfn[lca], 1);
				if (fa) bit.Upd(dfn[fa], 1);
				--cnt;
			}
			if (it.w > mid) vis[i] = 1;
			else _id[++L] = id[i], vis[i] = 0;
		}
		if (it.op == 2) {
			int res = bit.Qry(dfn[it.u], dfn[it.u] + sz[it.u] - 1);
			if (res == cnt) {
				_id[++L] = id[i], vis[i] = 0;
			} else {
				vis[i] = 1;
			}
		}
	}
	R = L;
	rep(i, ql, qr) 
		if (vis[i]) _id[++R] = id[i];
	rep(i, ql, qr) {
		auto it = Op[id[i]];
		if (it.op == 0 && it.w >= mid) {
			int u = it.u, v = it.v, lca = LCA(u, v), fa = Fa[lca];
			bit.Upd(dfn[u], -1), bit.Upd(dfn[v], -1), bit.Upd(dfn[lca], 1);
			if (fa) bit.Upd(dfn[fa], 1);
		}
		if (it.op == 1 && it.w >= mid) {
			int u = it.u, v = it.v, lca = LCA(u, v), fa = Fa[lca];
			bit.Upd(dfn[u], 1), bit.Upd(dfn[v], 1), bit.Upd(dfn[lca], -1);
			if (fa) bit.Upd(dfn[fa], -1);
		}
	}
	rep(i, ql, qr) id[i] = _id[i];
	Solve(l, mid, ql, L), Solve(mid + 1, r, L + 1, qr);
}
 
int main() {
	ios::sync_with_stdio(false), cin.tie(nullptr);
	cin >> n >> m;
	rep(i, 1, n - 1) {
		int u, v;
		cin >> u >> v, G[u].push_back(v), G[v].push_back(u);
	}
	int mx = 0;
	rep(i, 1, m) {
		int x;
		cin >> Op[i].op;
		if (Op[i].op == 0) 
			cin >> Op[i].u >> Op[i].v >> Op[i].w, mx = max(mx, Op[i].w);
		if (Op[i].op == 1) 
			cin >> x, Op[i].u = Op[x].u, Op[i].v = Op[x].v, Op[i].w = Op[x].w;
		if (Op[i].op == 2) 
			cin >> Op[i].u, Op[i].id = ++tot;
		id[i] = i;
	}
	DFS1(1, 0), DFS2(1, 1), Solve(1, mx + 1, 1, m);
	rep(i, 1, tot) 
		cout << ans[i] << '\n';
	return 0;
}

本文作者：Laijinyi

本文链接：https://www.cnblogs.com/laijinyi/p/18449769

版权声明：本作品采用知识共享署名-非商业性使用-禁止演绎 2.5 中国大陆许可协议进行许可。

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