这个复杂度感觉没什么问题啊为什么T了5个？

#include<cstdio>
#include<vector>
#define rt register int
using namespace std;
typedef long long ll;
const ll mod=998244353;
const int N=1e6;
int n,m;
ll a[N+1],tmul=1;
ll t[N+1];
ll add[N+1],pos[N+1],mul[N+1],f[N+1];
ll enda[N+1];
inline int read()
{
    char c=getchar();
    int f=1,s=0;
    while(c<48||c>57){if(c=='-')f=-f;c=getchar();}
    while(c>=48&&c<=57){s=(s<<3)+(s<<1)+(c^48);c=getchar();}
    return s*f;
}
int adj[N+1],nadj[N+1],nex[N+1],nnex[N+1],ter[N+1],nter[N+1],sum,nsum;
inline void adde(int x,int y){ter[++sum]=y,nex[sum]=adj[x],adj[x]=sum;}
inline void nadd(int x,int y){nter[++nsum]=y,nnex[nsum]=nadj[x],nadj[x]=nsum;}
inline void dfs(int x)
{
    for(rt i=nadj[x];i;i=nnex[i])
    {
        int now=nter[i];
        if(t[now]==1) enda[pos[now]]=(enda[pos[now]]+add[now]*tmul%mod)%mod;
        if(tmul!=0)if(t[now]==2) tmul=tmul*mul[now]%mod;
        if(t[now]==3)
        dfs(now);
    }   
}
int main()
{
    n=read();
    int c,sg,Q;
    for(rt i=1;i<=n;i++)
     a[i]=read();
    m=read();
    for(rt i=1;i<=m;i++)
     {
        t[i]=read();
        if(t[i]==1)
        pos[i]=read(),add[i]=read();
        if(t[i]==2)
        mul[i]=read();
        if(t[i]==3)
        {
         c=read();
         for(rt j=1;j<=c;j++)   
          {
            sg=read();
            nadd(i,sg);
           }
         }
     }
     Q=read();
     for(rt i=1;i<=Q;i++)
      f[i]=read();
     for(rt i=1;i<=Q;i++)
      adde(f[0],f[i]);
    for(rt i=adj[0];i;i=nex[i])
    {
        rt now=ter[i];
        if(t[now]==1) enda[pos[now]]=(enda[pos[now]]+add[now]*tmul%mod)%mod;
    if(tmul!=0)if(t[now]==2) tmul=tmul*mul[now]%mod;
        if(t[now]==3)
        dfs(now);
    } 
    for(rt i=1;i<=n;i++)
    printf("%lld ",(a[i]*tmul%mod+enda[i])%mod);
    return 0;
}