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[ZJOI2008]树的统计Count

 
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题目大意:

给出一棵树,点上有点权

有两种询问,询问两点之间树链上的点权和,询问两点之间树链上最大点权

点数<=30000,询问树<=200000


最裸的动态树,操作少,维护的值简单


{int y=fa(x),z=fa(y);if(lc(z)==y)lc(z)=x;else if(rc(z)==y)rc(z)=x;fa(x)=z;fa(rc(x))=y;lc(y)=rc(x);rc(x)=y;fa(y)=x;Update(y);}{int y=fa(x),z=fa(y);if(lc(z)==y)lc(z)=x;else if(rc(z)==y) rc(z)=x;fa(x)=z;fa(lc(x))=y;rc(y)=lc(x);lc(x)=y;fa(y)=x;Update(y);}void Splay(int x){for(int y,z;!isRoot(x);){y=fa(x);z=fa(y);if(isRoot(y))if(lc(y)==x)Zig(x);else Zag(x);elseif(lc(z)==y)if(lc(y)==x)Zig(y),Zig(x);else Zag(x),Zig(x);elseif(lc(y)==x)Zig(x),Zag(x);else Zag(y),Zag(x);}Update(x);}void DFS(int u,int f){key(u)=w[u];erep(i,e,h[u])if(e[i].node!=f)DFS(e[i].node,u),fa(e[i].node)=u;}void Init(){int a,b;scanf("%d",&n);rep(i,1,n-1){scanf("%d%d",&a,&b);add(a,b);add(b,a);}rep(i,1,n)scanf("%d",w+i);key(0)=Max(0)=-oo;DFS(1,0);}inline void Expose(int x){int y;for(y=0;x;x=fa(x)){Splay(x);rc(x)=y;Update(x);y=x;}}inline void Change(int x,int y){key(x)=y;Splay(x);}inline int Qmax(int x,int y){Expose(y);for(y=0;x;x=fa(x)){Splay(x);if(!fa(x))return max(key(x),max(Max(rc(x)),Max(y)));rc(x)=y;Update(x);y=x;}}inline int Qsum(int x,int y){Expose(y);for(y=0;x;x=fa(x)){Splay(x);if(!fa(x))return sum(rc(x))+key(x)+sum(y);rc(x)=y;Update(x);y=x;}}void Work(){char ch[10];int a,b;scanf("%d",&Q);rep(i,1,Q){scanf("%s%d%d",ch,&a,&b);if(ch[0]=='C')Change(a,b);else{if(ch[1]=='M')printf("%d\n",Qmax(a,b));else printf("%d\n",Qsum(a,b));}}}int main(){//freopen((name+in).c_str(),"r",stdin);//freopen((name+out).c_str(),"w",stdout);Init();Work();return 0;}




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