CF832 D LCA倍增 裸

有询问$a,b,c$，求a到c路径上，同时是a到b路径的点的个数。其中询问中的a,b,c可任意选择作为起点或终点，求一组询问中最大值。

LCA用于计算树上点对间距离，对于一组询问求深度最大的点作为起点，再在其中找最大距离的点就可以了

/** @Date    : 2017-08-04 20:17:42
  * @FileName: D LCA.cpp
  * @Platform: Windows
  * @Author  : Lweleth (SoungEarlf@gmail.com)
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <bits/stdc++.h>
#define LL long long
#define PII pair
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;

int n, q;
vectoredg[N];
int deg[N];
int fa[N][20];

void init()
{
	for(int j = 1; (1 << j) <= n; j++)
	{
		for(int i = 1; i <= n; i++)
		{
			fa[i][j] = fa[fa[i][j - 1]][j - 1];
		}
	}
}

void dfs(int x, int pre, int dep)
{
	deg[x] = dep;
	for(auto i: edg[x])
	{
		if(i == pre)
			continue;
		fa[i][0] = x;
		dfs(i, x, dep + 1);
	}
}

int lca(int a, int b)
{
	if(deg[a] > deg[b])
		swap(a, b);
	int det = deg[b] - deg[a];
	for(int i = 0; (1 << i) <= det; i++)
	{
		if((1 << i) & det)
			b = fa[b][i];
	}
	if(a == b)
		return a;
	for(int i = 19; i >= 0; i--)
	{
		if(fa[a][i] == fa[b][i])
			continue;
		a = fa[a][i];
		b = fa[b][i];
	}
	return fa[a][0];
}

int dis(int a, int b)
{
	return deg[a] + deg[b] - 2 * deg[lca(a, b)];
}

int main()
{
	while(cin >> n >> q)
	{
		for(int i = 2; i <= n; i++)
		{
			int x;
			scanf("%d", &x);
			edg[x].PB(i);
			edg[i].PB(x);
		}
		dfs(1, -1, 0);
		init();
		while(q--)
		{
			int x, y, z;
			scanf("%d%d%d", &x, &y, &z);
			int f1 = lca(x, y);
			int f2 = lca(x, z);
			int f3 = lca(y, z);
			if(deg[f1] < deg[f2]) f1 = f2;
			if(deg[f1] < deg[f3]) f1 = f3;
			int ans = max(dis(f1, x), max(dis(f1, y), dis(f1, z)));
			printf("%d\n", ans + 1);
		}
	}
    return 0;
}