原来左偏树还可以打tag，get了
和线段树打tag一样，时不时Push_Down就好了
然后这里显然也是要先乘法后加法的
tag打上了之后还是其他一般左偏树差不多，有些细节注意一下
然后开 long long!!!
1 #include<bits/stdc++.h> 2 #define int long long 3 #define writeln(x) write(x),puts("") 4 #define writep(x) write(x),putchar(' ') 5 using namespace std; 6 inline int read(){ 7 int ans=0,f=1;char chr=getchar(); 8 while(!isdigit(chr)){if(chr=='-') f=-1;chr=getchar();} 9 while(isdigit(chr)){ans=(ans<<3)+(ans<<1)+chr-48;chr=getchar();} 10 return ans*f; 11 }void write(int x){ 12 if(x<0) putchar('-'),x=-x; 13 if(x>9) write(x/10); 14 putchar(x%10+'0'); 15 }const int M = 3e5+5; 16 int a[M],h[M],v[M],c[M],n,m,head[M],ver[M<<1],nxt[M<<1],tot,val[M],son[M][2],dis[M],fa[M]; 17 int mul[M],add[M],dep[M],ans1[M],ans2[M]; 18 inline void Add(int x,int y){ver[++tot]=y;nxt[tot]=head[x];head[x]=tot;} 19 #define ls son[x][0] 20 #define rs son[x][1] 21 inline void Down(int x,int mull,int addd){ 22 if(!x) return; 23 val[x]*=mull,val[x]+=addd; 24 mul[x]*=mull,add[x]*=mull,add[x]+=addd; 25 } 26 inline void Push_Down(int x){ 27 Down(ls,mul[x],add[x]); 28 Down(rs,mul[x],add[x]); 29 mul[x]=1,add[x]=0; 30 } 31 int Merge(int x,int y){ 32 if(!x||!y) return x|y; 33 Push_Down(x);Push_Down(y); 34 if(val[x]>val[y]) swap(x,y); 35 rs=Merge(rs,y); 36 if(dis[ls]<dis[rs]) swap(ls,rs); 37 dis[x]=dis[rs]+1; 38 return x; 39 }int Pop(int x){return Merge(ls,rs);} 40 void dfs(int x){ 41 for(int i=head[x];i;i=nxt[i]){ 42 dep[ver[i]]=dep[x]+1; 43 dfs(ver[i]); 44 fa[x]=Merge(fa[x],fa[ver[i]]); 45 } 46 while(fa[x]&&val[fa[x]]<h[x]){ 47 Push_Down(fa[x]); 48 ans1[x]++;ans2[fa[x]]=dep[c[fa[x]]]-dep[x]; 49 fa[x]=Pop(fa[x]); 50 }if(a[x]) Down(fa[x],v[x],0); 51 else Down(fa[x],1,v[x]); 52 } 53 signed main(){ 54 n=read(),m=read(); 55 for(int i=1;i<=n;i++) h[i]=read(); 56 for(int x,i=2;i<=n;i++) x=read(),Add(x,i),a[i]=read(),v[i]=read(); 57 for(int i=1;i<=m;i++){ 58 val[i]=read(),c[i]=read(),mul[i]=1; 59 fa[c[i]]=Merge(fa[c[i]],i); 60 }dep[1]=1,dfs(1); 61 while(fa[1]){ 62 Push_Down(fa[1]); 63 ans2[fa[1]]=dep[c[fa[1]]]; 64 fa[1]=Pop(fa[1]); 65 } 66 for(int i=1;i<=n;i++) writeln(ans1[i]); 67 for(int i=1;i<=n;i++) writeln(ans2[i]); 68 return 0; 69 }