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;
}