一开始写的暴力合并 卡n^2过的不是正解

看正解是类似 虚拟点+树形DP的思路 很巧妙 记录一下

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int,int>;
#define int long long
const int N = 3e5+10;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;
int gcd(int a,int b){return b?a:gcd(b,a%b);}
int lcm(int a,int b){return a*b/gcd(a,b);}
int qmi(int a,int b,int mod){int res=1;while(b){if(b&1)res=res*a%mod;b>>=1;a=a*a%mod;}return res;}


int n,q,m;
int p[N];
int dp[N];
int find(int x){
	if(x!=p[x])p[x] = find(p[x]);
	return p[x];
}
int e[N],ne[N],w[N],h[N],idx;
void add(int a,int b,int c=0){
	e[idx] = b,ne[idx] = h[a],w[idx] = c,h[a] = idx++;
}


void dfs(int u,int fa){
	dp[u]+=dp[fa];
	for(int i=h[u];~i;i=ne[i]){
		int j = e[i];
		dfs(j,u);
	}
}


void solve()
{
	cin>>n>>m;
	int root = n+1;
	memset(h,-1,sizeof h);
	for(int i=1;i<=2*n+10;++i)p[i] = i;
	while(m--){
		int op,a,b;cin>>op>>a>>b;
		if(op==1){
			int pa = find(a),pb = find(b);
			if(pa==pb)continue;
			p[pa] = root;
			p[pb] = root;
			add(root,pa);
			add(root,pb);
			++root;
		}else{
			dp[find(a)]+=b;
		}
	}
	
	for(int i=n+1;i<root;++i)if(find(i)==i)dfs(i,0);
	for(int i=1;i<=n;i++)cout<<dp[i]<<" ";
	
	

}

signed main()
{
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	int _;
	//cin>>_;
	_ = 1;
	while(_--)solve();
	return 0;
}