BZOJ - 1036 树的统计Count (LCT)

LCT试炼题（代码量居然完爆树剖？）

 #include<bits/stdc++.h>
 using namespace std;
 const int N=3e4+,inf=0x3f3f3f3f;
 int fa[N],ch[N][],flp[N],n,m,sta[N],tp,val[N],sum[N],mx[N],tu[N],tv[N];
 bool isrt(int u) {return ch[fa[u]][]!=u&&ch[fa[u]][]!=u;}
 int sf(int u) {return ch[fa[u]][]==u;}
 void pu(int u) {sum[u]=sum[ch[u][]]+sum[ch[u][]]+val[u],mx[u]=max(max(mx[ch[u][]],mx[ch[u][]]),val[u]);}
 void pd(int u) {if(flp[u])flp[u]=,flp[ch[u][]]^=,flp[ch[u][]]^=,swap(ch[u][],ch[u][]);}
 void rot(int u) {
     int v=fa[u],w=fa[v],f=sf(u);
     if(!isrt(v))ch[w][sf(v)]=u;
     ch[v][f]=ch[u][f^],fa[ch[v][f]]=v;
     ch[u][f^]=v,fa[u]=fa[v],fa[v]=u;
     pu(v);
 }
 void splay(int u) {
     sta[tp=]=u;
     for(int v=u; !isrt(v); v=fa[v])sta[++tp]=fa[v];
     for(; ~tp; pd(sta[tp--]));
     for(; !isrt(u); rot(u))
         if(!isrt(fa[u])&&sf(fa[u])==sf(u))rot(fa[u]);
     pu(u);
 }
 void access(int u) {for(int v=; u; u=fa[v=u])splay(u),ch[u][]=v,pu(u);}
 void makert(int u) {access(u),splay(u),flp[u]^=;}
 void link(int u,int v) {makert(u),fa[u]=v;}
 void split(int u,int v) {makert(u),access(v),splay(v);}
 int main() {
     sum[]=,mx[]=~inf;
     scanf("%d",&n);
     for(int i=; i<n; ++i)scanf("%d%d",&tu[i],&tv[i]);
     for(int i=; i<=n; ++i)scanf("%d",&val[i]),mx[i]=sum[i]=val[i];
     for(int i=; i<n; ++i)link(tu[i],tv[i]);
     scanf("%d",&m);
     char s[];
     while(m--) {
         int x,y;
         scanf("%s%d%d",s,&x,&y);
         if(s[]=='S')split(x,y),printf("%d\n",sum[y]);
         else if(s[]=='M')split(x,y),printf("%d\n",mx[y]);
         else if(s[]=='H')splay(x),val[x]=y,pu(x);
     }
     return ;
 }