This problem is very small space, that O (nlogn) lca tree algorithm is here not useful, and can be considered a tree block.
This question block tree is based on the depth, i.e. the depth of each press \ (\ sqrt n \) points a, and then jump up a piece, has at lca jump.
For this problem, there are several such practices:
Consider selecting a number of key points in the tree, each seeking lca to jump up at the critical point recently, and then a key point of a jump up, jump up at the same critical point. As for the selection of key points, you can do this:
From light to dark turn from every point of view, if the jump up 2S-1 nodes are not the key point, it is set to S-Class ancestor key points. From the first key thus key point number strictly less than n / S, while each keypoint to its father direction equal to S. And jump up from each point, a key point to the first number of steps required <2S.
Of course, the tree may be random \ (\ sqrt n \) key points, such complexity is desirable to \ (O (\ sqrt n) \) a.
Another approach gods psj provided is recorded at each point jump up \ (\ sqrt n \) times the ancestors, and then each time the query first two points before the jump to the same depth (depth can jump \ (O (\ sqrt n) \) queries get, jump when the dynamic maintenance), then multiply like lca that can jump up to jump up, but the step is \ (\ sqrt n \) , and finally in step by step jump lca Department. This approach is twice the space above practices have space to card a card too.
I write here is the first approach, the details can refer to the code:
#include<cstdio>
#include<bitset>
#include<algorithm>
using namespace std;
#define N 900007
#define M 5007
const int t=300;
int fa[N],a[M],top[M],dep[M],cnt;
bitset<N> tag;
void mark(int x)
{
int v=fa[x],i;
for(i=1;i<2*t&&!tag[v];i++)
v=fa[v];
if(i==2*t)
{
for(i=1;i<=t;i++)
x=fa[x];
a[++cnt]=x;
tag[x]=1;
}
}
int getpos(int x)
{
int p=lower_bound(a+1,a+cnt+1,x)-a;
return p;
}
int find(int x)
{
while(!tag[x])x=fa[x];
return x;
}
int getdep(int x,int y)
{
int d=0;
while(x!=y)
d++,x=fa[x];
return d;
}
int getkth(int x,int k)
{
int i;
for(i=1;i<=k;i++)
x=fa[x];
return x;
}
int main()
{
int n,m,i,x,y;
scanf("%d%d",&n,&m);
for(i=2;i<=n;i++)
{
scanf("%d",&x);
fa[i]=x;
}
tag[1]=1;
a[++cnt]=1;
for(i=2;i<=n;i++)
mark(i);
sort(a+1,a+cnt+1);
dep[1]=1;
for(i=2;i<=cnt;i++)
{
top[i]=getpos(find(fa[a[i]]));
dep[i]=dep[top[i]]+1;
}
for(i=1;i<=m;i++)
{
scanf("%d%d",&x,&y);
int u=find(x),v=find(y);
u=getpos(u),v=getpos(v);
if(dep[v]>dep[u])swap(x,y),swap(u,v);
while(dep[u]>dep[v])
{
x=a[u];
u=top[u];
}
while(u!=v)
{
x=a[u],u=top[u];
y=a[v],v=top[v];
}
int dx=getdep(x,a[u]),dy=getdep(y,a[v]);
if(dy>dx)swap(x,y),swap(dx,dy);
x=getkth(x,dx-dy);
while(x!=y)
x=fa[x],y=fa[y];
printf("%d\n",x);
}
return 0;
}