P9638 「yyOI R1」youyou 的军训 题解

Kruskal 重构树板子。

考虑建边权从大到小排序后跑 Kruskal 的重构树，对于当前点 $u$ 和参数 $w$，只能走 $\geq w$ 边，即能去的点是 $u$ 祖先中最浅的点权 $\geq w$ 的点所在子树内每个叶子节点。

发现边权修改后，相对大小不变，直接在重构树点权修改即可。

倍增处理，复杂度 $O(q \log n)$。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <map>
using namespace std;

const int N = 8e5 + 5;
int p[N];

struct Edge
{
	int u, v, w;

	Edge() = default;

	Edge(int u, int v, int w)
		: u(u), v(v), w(w)
	{
	}

	bool operator==(const Edge& other) const
	{
		return u == other.u && v == other.v && w == other.w;
	}
}e[N];

int n, m, q;
int u[N], v[N], w[N];
int res[N];
int nfa[N][21];
map<pair<int, int>, int> mp;

class Union_Find
{
public:
	int fa[N], idx;
	int val[N], res[N];
	vector<int> G[N];
	void Init()
	{
		mp.clear();
		for (int i = 0; i < N; i++) fa[i] = i, val[i] = 0, G[i].clear();
		for (int i = 1; i <= n; i++) res[i] = 1;
		idx = n;
	}
	int find(int u)
	{
		return (fa[u] == u ? u : fa[u] = find(fa[u]));
	}
	void merge(int u, int v, int w)
	{
		if (find(u) == find(v)) return;
		mp[make_pair(u, v)] = ++idx;
		u = find(u), v = find(v);
		val[idx] = w;
		G[idx].emplace_back(u), G[idx].emplace_back(v);
		nfa[u][0] = nfa[v][0] = idx;
		fa[u] = fa[v] = idx;
		res[idx] += res[u] + res[v];
	}
}uf;

int val[N];
vector<int> G[N];

void kruskal()
{
	uf.Init();
	sort(e + 1, e + m + 1, [&](const Edge& x, const Edge& y) {return x.w > y.w; });
	for (int i = 1; i <= m; i++) uf.merge(e[i].u, e[i].v, e[i].w);
	for (int i = 0; i < N; i++) val[i] = uf.val[i], G[i].swap(uf.G[i]);
}

int get(int u, int x)
{
	for (int i = 20; i >= 0; i--)
	{
		if (nfa[u][i] && val[nfa[u][i]] >= x) u = nfa[u][i];
	}
	return u;
}

int main()
{
	ios::sync_with_stdio(0), cin.tie(0);
	cin >> n >> m >> q;
	for (int i = 1; i <= m; i++)
	{
		cin >> u[i] >> v[i] >> w[i];
		e[i] = Edge(u[i], v[i], w[i]);
	}
	kruskal();
	for (int i = 1; i <= 20; i++)
	{
		for (int j = 0; j < N; j++) nfa[j][i] = nfa[nfa[j][i - 1]][i - 1];
	}
	for (int i = 0; i < N; i++) res[i] = uf.res[i];
	int now = 0;
	while (q--)
	{
		int op;
		cin >> op;
		if (op == 1)
		{
			cin >> now;
		}
		else if (op == 2)
		{
			int x;
			cin >> x;
			cout << res[get(x, now)] << "\n";
		}
		else
		{
			int x, y;
			cin >> x >> y;
			if (mp.count(make_pair(u[x], v[x])))
			{
				val[mp[make_pair(u[x], v[x])]] = y;
			}
		}
	}
	return 0;
}