P3250 网络 题解

Solution

单次二分：问“重要度 \(\ge x\) 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 \(\ge mid+1\) 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 \(O(n\log^2 n)\) 的.

Code 1

\(\text{vector}\) 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;
}