【Topic】Double on the tree

Tarjan LCA

http://blog.csdn.net/cdy1206473601/article/details/77104910

explain

I wrote Tarjan LCA last time, but when the tree is a chain, the time complexity is very high! So, we can solve it with multiplication. However, under random data, Tarjan is faster than doubling!
Well, the idea of ​​doubling is to set a fa[i][j] to represent the first node of i$2^j$ancestors. Then you can use this fa array by solving upwards to achieve the multiplication effect.
Forming this fa array is done with a dfs. Transfer equation fa[i][j]=fa[fa[i][j-1]][j-1];

void dfs(int x)
{
    fa[x][0]=f[x];
    for (int i=1;i<=20;i++)
    {
        if (fa[x][i-1]<=1) break;
        fa[x][i]=fa[fa[x][i-1]][i-1]；
    }
    int i=last[x],yy=0;
    vis[x]=true;
    while (i)
    {
        yy=tov[i];
        if (vis[yy]==false)
        {
            f[yy]=x;
            deep[yy]=deep[x]+1;
            dfs(yy);
        }
        i=next[i];
    }
}

Then it's doubling, first looking for the depth of these two points, until the depths of the two points are equal~

int lca_len(int x,int y)
{
    if (deep[x]<deep[y]) 
    {
        int t=x;
        x=y;
        y=t;
    }
    for (int i=20;i>=0;i--)//使其深度相等。
    {
        if (fa[x][i]!=0&&deep[y]<=deep[fa[x][i]]) 
            x=fa[x][i];
        if (x==y) break;
    }
    if (x==y) return x;
    for (int i=20;i>=0;i--)
    {
        if (fa[x][i]!=fa[y][i]) 
        {
            x=fa[x][i];
            y=fa[y][i];
        }
        if (x==y) break;
    }
    if (f[x]==f[y])
    {
        x=fa[x][0];
        y=fa[y][0]; 
    }
    return x;//最后Lca就是它们本身啦。
}