Black Technology - tree turn a dfs dfs sectional twice!

Black Technology - tree cross twice \ (dfs \) turn a \ (dfs \) !

Weight is well known that the tree is usually split chain to \ (dfs \) twice, like this:

int Fa[N],dep[N],Sz[N],son[N];
void dfs1(int x,int pre){
    Fa[x]=pre,dep[x]=dep[pre]+1;
    Sz[x]=1;
    erep(i,G,x){
        int y=G.to[i];
        if(y==pre)continue;
        dfs(y,x);
        Sz[x]+=Sz[y];
        (Sz[y]>Sz[son[x]])&&(son[x]=y);
    }
}
int L[N],R[N],Id,top[N];
void dfs2(int x,int tp){
    top[x]=tp;
    if(son[x])dfs2(son[x],tp);
    erep(i,G,x){
        int y=G.to[i];
        if(y==Fa[x]||y==son[x])continue;
        dfs2(y,y);
    }
}

But in some \ (n \) is relatively large and theOften cancer card titleWe used twice if \ (dfs \) is likely to occur \ (tle \) situations.

In this case, the nature of the case, if we can not optimize algorithms we often need card.

How card it? Use \ (dfs \) sequence:

code show as below:

int Fa[N],dep[N],Sz[N],son[N],L[N],R[N],Id[N],cnt,top[N];
void dfs1(int x,int pre){
    Fa[x]=pre,dep[x]=dep[pre]+1;
    Sz[x]=1,L[x]=++cnt,Id[cnt]=x;
    erep(i,G,x){
        int y=G.to[i];
        if(y==pre)continue;
        dfs(y,x);
        Sz[x]+=Sz[y];
        (Sz[y]>Sz[son[x]])&&(son[x]=y);
    }
    R[x]=cnt;
}
rep(i,1,n)top[Id[i]]=Id[i]==son[Fa[Id[i]]]?top[Fa[Id[i]]]:Id[i];

So we only used once \ (dfs \) to complete the pre-operating section of the tree.

Do not underestimate this a recursive, he \ (\ n) is large, you can soon probably \ (1 \) times.