LOJ6066：「2017 山东一轮集训 Day3」第二题

传送门
二分答案 (k)，考虑如何 (hash) 使得做起来方便
把每个点挂在 (k+1) 级祖先上，考虑在祖先上删除
这道题巧妙在于其可以对于 (dfs) 序/括号序列 (hash)
这样在 (k+1) 级祖先上暴力删除就好了

# include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;

const int maxn(2e5 + 5);
const ull base(19260817);

unordered_map <ull, int> hash_table;
int first[maxn], cnt, n, idx, dfn[maxn], ed[maxn], mxd[maxn], st[maxn], top;
ull val[maxn], pw[maxn], hsh[maxn];
vector <int> son[maxn], kson[maxn];

inline void Dfs1(int u) {
	dfn[u] = ++idx, val[idx] = 233;
	for (auto v : son[u]) Dfs1(v), mxd[u] = max(mxd[u], mxd[v]);
	ed[u] = ++idx, ++mxd[u], val[idx] = 131;
}

inline void Dfs2(int u, int k) {
	st[++top] = u;
	if (top - 1 > k) kson[st[top - k - 1]].push_back(u);
	for (auto v : son[u]) Dfs2(v, k);
	--top;
}

inline ull Hash(int l, int r) {
	return hsh[r] - hsh[l - 1] * pw[r - l + 1];
}

inline int Calc(int k) {
	register int i, l, r;
	register ull v;
	hash_table.clear();
	for (i = 1; i <= n; ++i) kson[i].clear();
	Dfs2(1, k);
	for (i = 1; i <= n; ++i)
		if (mxd[i] > k) {
			l = dfn[i], v = 0;
			for (auto to : kson[i]) {
				r = dfn[to] - 1;
				v = v * pw[r - l + 1] + Hash(l, r);
				l = ed[to] + 1;
			}
			r = ed[i], v = v * pw[r - l + 1] + Hash(l, r);
			if (hash_table.count(v)) return 1;
			hash_table[v] = 1;
		}
	return 0;
}

int main() {
	register int i, v, x, l, r, mid, ans;
	scanf("%d", &n);
	for (i = 1; i <= n; ++i)
		for (scanf("%d", &x); x; --x) scanf("%d", &v), son[i].push_back(v);
	Dfs1(1), pw[0] = 1;
	for (i = 1; i <= idx; ++i) pw[i] = pw[i - 1] * base;
	for (i = 1; i <= idx; ++i) hsh[i] = hsh[i - 1] * base + val[i];
	l = 2, r = n, ans = 1;
	while (l <= r) Calc(mid = (l + r) >> 1) ? ans = mid, l = mid + 1 : r = mid - 1;
	printf("%d
", ans);
	return 0;
}