「LOJ6066」「2017 山东一轮集训 Day3」第二题-树hash

Description

链接

考虑对括号序列进行哈希。

二分答案 $k$。考虑一个点的 $k-$子树就是它的子树去掉距离它大于 $k$的节点，对应到括号序列上就是一个大区间减去若干个小区间。把每个点挂在它的 $k+1$级祖先上，然后计算即可。

Solution

考虑对括号序列进行哈希。

二分答案 $k$。考虑一个点的 $k-$子树就是它的子树去掉距离它大于 $k$的节点，对应到括号序列上就是一个大区间减去若干个小区间。把每个点挂在它的 $k+1$级祖先上，然后计算即可。

#include <bits/stdc++.h>
using namespace std;

typedef long long lint;
const int mod[2] = {998244353, 1004535809}, base[2] = {5007449, 5669};
const int maxn = 100005;

int n, son[maxn];

struct edge
{
	int to, next;
} e[maxn * 2];
int h[maxn], tot;

int dfn[maxn], low[maxn], s[maxn * 2], len[maxn], Time;
int pw[maxn * 2][2], hsh[maxn * 2][2];
int stk[maxn], top;
vector<int> vec[maxn];
unordered_map<int, int> g[2];

inline int gi()
{
	char c = getchar();
	while (c < '0' || c > '9') c = getchar();
	int sum = 0;
	while ('0' <= c && c <= '9') sum = sum * 10 + c - 48, c = getchar();
	return sum;
}

inline void add(int u, int v)
{
	e[++tot] = (edge) {v, h[u]};
	h[u] = tot;
}

void dfs(int u)
{
	s[dfn[u] = ++Time] = 1;
	for (int i = h[u], v; v = e[i].to, i; i = e[i].next)
		dfs(v), len[u] = max(len[u], len[v]);
	++len[u]; low[u] = ++Time;
}

void dfs(int u, int k)
{
	stk[++top] = u;
	if (top > k + 1) vec[stk[top - k - 1]].push_back(u);
	for (int i = h[u], v; v = e[i].to, i; i = e[i].next)
		dfs(v, k);
	--top;
}

void add(int *val, int l, int r)
{
	val[0] = (hsh[r][0] + (lint)(val[0] - hsh[l - 1][0] + mod[0]) * pw[r - l + 1][0]) % mod[0];
	val[1] = (hsh[r][1] + (lint)(val[1] - hsh[l - 1][1] + mod[1]) * pw[r - l + 1][1]) % mod[1];
}

bool check(int k)
{
	for (int i = 1; i <= n; ++i) vec[i].clear();
	dfs(1, k);
	g[0].clear(); g[1].clear();
	for (int i = 1; i <= n; ++i) {
		if (len[i] <= k) continue;
		int val[2] = {0, 0}, l = dfn[i], r;
		for (int v : vec[i]) {
			r = dfn[v] - 1;
			if (l <= r) add(val, l, r);
			l = low[v] + 1;
		}
		add(val, l, low[i]);
		if (g[0].count(val[0]) && g[1].count(val[1])) return 1;
		g[0][val[0]] = 1; g[1][val[1]] = 1;
	}
	return 0;
}

int main()
{
	n = gi();
	for (int x, i = 1; i <= n; ++i) {
		x = gi();
		for (int j = 1; j <= x; ++j) son[j] = gi();
		for (int j = x; j >= 1; --j) add(i, son[j]);
	}

	dfs(1);

	pw[0][0] = pw[0][1] = 1;
	for (int i = 1; i <= Time; ++i) {
		for (int j = 0; j < 2; ++j) {
			pw[i][j] = (lint)pw[i - 1][j] * base[j] % mod[j];
			hsh[i][j] = ((lint)hsh[i - 1][j] * base[j] + (s[i] ? 19 : 23)) % mod[j];
		}
	}

	int l = 0, r = len[1], mid;
	while (l < r) {
		mid = (l + r + 1) >> 1;
		if (check(mid)) l = mid;
		else r = mid - 1;
	}

	printf("%d\n", l);
	
	
	return 0;
}