D. Misha, Grisha and Underground(LCA+求解重叠路径)

$考虑给定a,b,c,其中c为终点,如何求ac和bc的重叠路径呢?$

$令dis(q,w)表示q和w的距离$

$那么重叠路径是(\ dis(a,c)+dis(b,c)-dis(a,b)\ )/2$

$假如a和b分别在c的两侧,显然dis(a,c)+dis(b,c)=dis(a,b),满足$

$若在同一侧,dis(a,b)表示的是到公共点前的路径$

$减去除以2即可$

$求dis用lca来实现$

#include <bits/stdc++.h>
using namespace std;
const int maxn=3e5+10;
struct p{
	int to,nxt;
}d[maxn]; int head[maxn],cnt=1,n,q;
void add(int u,int v){
	d[cnt]=(p){ v,head[u] },head[u]=cnt++;
}
int fa[maxn][22],deep[maxn],lg[maxn];
void dfs(int u,int father)
{
	deep[u]=deep[father]+1,fa[u][0]=father;
	for(int i=1;i<=lg[ deep[u] ];i++)
		fa[u][i]=fa[ fa[u][i-1] ][i-1];
	for(int i=head[u];i;i=d[i].nxt)
	{
		int v=d[i].to;
		if( v==father )	continue;
		dfs(v,u);
	}
}

int lca(int x,int y)
{
	if( deep[x]<deep[y] )	swap(x,y);
	while( deep[x]>deep[y] )
		x=fa[x][ lg[deep[x]-deep[y]]-1 ];
	if( x==y )	return x;
	for(int i=lg[ deep[x] ]-1;i>=0;i--)
		if( fa[x][i]!=fa[y][i] )
			x=fa[x][i],y=fa[y][i];
	return fa[x][0];
}
void init(){
	for(int i=1;i<=200000;i++)	
		lg[i]=lg[i-1]+(i==(1<<lg[i-1]) );
}
int get_dis(int x,int y){
	int u=lca(x,y);
//	cout << x << "和" << y << "的lca是" << u << endl;
	return deep[x]+deep[y]-2*deep[u];
} 
int main()
{
	cin >> n >> q;
	init();
	for(int i=2,l;i<=n;i++)
	{
		scanf("%d",&l);
		add(i,l); add(l,i);
	}
	dfs(1,0);
	while( q-- )
	{
		int a,b,c,ans;
		scanf("%d%d%d",&a,&b,&c);
		int ab=get_dis(a,b);
		int bc=get_dis(b,c);
		int ac=get_dis(a,c);
	//	cout << ab << " " << bc << " " << ac << endl;
		int na=( ac+ab-bc )/2;
		int nb=( ab+bc-ac )/2;
		int nc=( bc+ac-ab )/2;
		ans=max(na,max(nb,nc) )+1;
		printf("%d\n",ans);
	}
}