P8827 [传智杯 #3 初赛] 森林 题解

考虑操作有删边但没有恢复边的操作，很容易想到离线后逆序操作的套路。

逆序后操作即为只有加边没有删边了，并查集维护即可。

但是注意到题目要求点权和，所以带权并查集维护，修改权值也可以轻松做，即在父亲的点上减去本身的贡献加上新的贡献即可。

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

const int N = 1e5 + 5;

int ans[N], n, m;
int a[N];
bool v[N];

struct Query
{
	int op, u, val, id;
}q[N];

class Union_Find
{
public:
	int fa[N], val[N];
	void build()
	{
		for (int i = 0; i < N; i++) fa[i] = i, val[i] = a[i];
	}
	int find(int u)
	{
		return fa[u] == u ? u : fa[u] = find(fa[u]);
	}
	void merge(int u, int v)
	{
		int x = find(u), y = find(v);
		fa[x] = y;
		val[y] += val[x];
	}
}f;

pair<int, int> EDGE[N];
int idx, lst[N];

int main()
{
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
	for (int i = 1; i < n; i++)
	{
		int u, v;
		scanf("%d%d", &u, &v);
		EDGE[i] = make_pair(u, v);
	}
	for (int i = 1; i <= m; i++)
	{
		int op;
		scanf("%d", &op);
		q[i].op = op;
		if (op == 1)
		{
			scanf("%d", &op);
			v[op] = 1;
			q[i].u = op;
		}
		else if (op == 2)
		{
			int x, y;
			scanf("%d%d", &x, &y);
			q[i].u = x;
			q[i].val = y;
			lst[i] = a[x];
			a[x] = y;
		}
		else
		{
			scanf("%d", &op);
			q[i].id = ++idx;
			q[i].u = op;
		}
	}
	f.build();
	for (int i = 1; i < n; i++)
	{
		if (!v[i])
		{
			f.merge(EDGE[i].first, EDGE[i].second);
		}
	}
	for (int i = m; i >= 1; i--)
	{
		int op = q[i].op;
		if (op == 1)
		{
			int u = q[i].u;
			f.merge(EDGE[u].first, EDGE[u].second);
		}
		else if (op == 2)
		{
			int u = q[i].u, v = q[i].val;
			int ff = f.find(u);
			f.val[ff] -= v;
			f.val[ff] += lst[i];
			//a[u] = v;
		}
		else
		{
			int u = q[i].u;
			ans[q[i].id] = f.val[f.find(u)];
		}
	}
	for (int i = 1; i <= idx; i++) printf("%d\n", ans[i]);
	return 0;
}