【LG3250】[HNOI2016]网络

【LG3250】[HNOI2016]网络

题面

洛谷

题解

30pts

对于\(m\leq 2000\)，直接判断一下这个个点是否断掉一个交互，没断掉的里面取\(max\)即可，复杂度\(O(m^2\log n)\)。

另20pts

对于无删除操作的，用线段树维护，

我们将一条路径的补集全部打上那条路径重要度的标记，这样我们断一条边时直接单点查询即可。

据说这样子再改改可以变成一种树剖加堆的做法，但是好像现在在bzoj被卡了

100pts

想到二分答案，若所有权值\(\geq mid\)的路径都过\(x\)，那么\(ans<mid\)，反之若有不过\(x\)的，则说明\(ans\geq mid\)。

这个可以用树套树维护，因为树状数组套线段树会炸空间，所以就线段树套平衡树，复杂度\(O(m\log ^3n)\)。

考虑整体二分，那么我们按照套路按时间加入，将一条路径树上差分，

碰到一次查询直接看经过它的路径条数即可，复杂度\(O(m\log^2 n)\)。

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
inline int gi() {
    register int data = 0, w = 1;
    register char ch = 0;
    while (!isdigit(ch) && ch != '-') ch = getchar();
    if (ch == '-') w = -1, ch = getchar();
    while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();
    return w * data;
}
const int MAX_N = 1e5 + 5, MAX_M = 2e5 + 5;
struct Graph { int to, next; } e[MAX_N << 1]; int fir[MAX_N], e_cnt;
void clearGraph() { memset(fir, -1, sizeof(fir)); e_cnt = 0; }
void Add_Edge(int u, int v) { e[e_cnt] = (Graph){v, fir[u]}; fir[u] = e_cnt++; }
int dep[MAX_N], fa[MAX_N], size[MAX_N], top[MAX_N], son[MAX_N];
int L[MAX_N], R[MAX_N], tim;
void dfs1(int x) {
	dep[x] = dep[fa[x]] + 1, size[x] = 1;
	for (int i = fir[x]; ~i; i = e[i].next) {
		int v = e[i].to; if (v == fa[x]) continue;
		fa[v] = x, dfs1(v), size[x] += size[v];
		if (size[son[x]] < size[v]) son[x] = v;
	}
}
void dfs2(int x, int tp) {
	top[x] = tp, L[x] = ++tim;
	if (son[x]) dfs2(son[x], tp);
	for (int i = fir[x]; ~i; i = e[i].next) {
		int v = e[i].to; if (v == son[x] || v == fa[x]) continue;
		dfs2(v, v);
	}
	R[x] = tim;
}
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;
}
int N, M;
struct Option { int op, a, b, lca, v, id; } q[MAX_M], lq[MAX_M], rq[MAX_M];
int ans[MAX_M];
int c[MAX_N];
inline int lb(int x) { return x & -x; }
int sum(int x) { int res = 0; while (x) res += c[x], x -= lb(x); return res; }
void add(int x, int v) { while (x <= N) c[x] += v, x += lb(x); }
void Div(int lval, int rval, int st, int ed) {
	if (st > ed) return ;
	if (lval == rval) {
		for (int i = st; i <= ed; i++)
			if (q[i].op == 2) ans[q[i].id] = lval;
		return ;
	}
	int mid = (lval + rval) >> 1, lt = 0, rt = 0, cnt = 0;
	for (int i = st; i <= ed; i++) {
		if (q[i].op != 2) {
			if (q[i].v > mid) {
				int v = q[i].op ? -1 : 1;
				cnt += v;
				add(L[q[i].a], v), add(L[q[i].b], v), add(L[q[i].lca], -v);
				if (fa[q[i].lca]) add(L[fa[q[i].lca]], -v);
				rq[++rt] = q[i];
			} else lq[++lt] = q[i];
		} else {
			int res = sum(R[q[i].a]) - sum(L[q[i].a] - 1);
			if (cnt <= res) lq[++lt] = q[i];
			else rq[++rt] = q[i];
		}
	}
	for (int i = 1; i <= rt; i++) {
		if (rq[i].op != 2) {
			if (rq[i].v > mid) {
				int v = rq[i].op ? 1 : -1;
				add(L[rq[i].a], v), add(L[rq[i].b], v), add(L[rq[i].lca], -v);
				if (fa[rq[i].lca]) add(L[fa[rq[i].lca]], -v);
			}
		}
	}
	for (int i = 1; i <= lt; i++) q[st + i - 1] = lq[i];
	for (int i = 1; i <= rt; i++) q[st + lt - 1 + i] = rq[i];
	Div(lval, mid, st, st + lt - 1);
	Div(mid + 1, rval, st + lt, ed);
}
int h[MAX_N], cnt;
int main () {
#ifndef ONLINE_JUDGE
    freopen("cpp.in", "r", stdin);
#endif
	clearGraph();
	N = gi(), M = gi();
	for (int i = 1; i < N; i++) {
		int u = gi(), v = gi();
		Add_Edge(u, v), Add_Edge(v, u);
	}
	dfs1(1), dfs2(1, 1);
	int q_cnt = 0;
	for (int i = 1; i <= M; i++) {
		int op = gi();
		if (op == 0) {
			int a = gi(), b = gi();
		    q[i] = (Option){op, a, b, LCA(a, b), gi(), 0};
			h[++cnt] = q[i].v;
		}
		if (op == 1) q[i] = q[gi()], q[i].op = 1;
		if (op == 2) q[i] = (Option){op, gi(), 0, 0, 0, ++q_cnt};
	}
	h[++cnt] = -1;
	sort(&h[1], &h[cnt + 1]); cnt = unique(&h[1], &h[cnt + 1]) - h - 1;
	for (int i = 1; i <= M; i++) if (q[i].op != 2) q[i].v = lower_bound(&h[1], &h[cnt + 1], q[i].v) - h;
	Div(1, cnt, 1, M);
	for (int i = 1; i <= q_cnt; i++) printf("%d\n", h[ans[i]]);
    return 0;
}