bzoj1787.紧急集合(倍增LCA)

有一棵树有 N 个结点，共有 M 次查询，每次询问树上三个点到树上哪个点汇合距离最近，输出这个汇合点和最短距离

倍增 LCA，每次询问两两求 LCA ，三个 LCA 深度最深的点是集合点。

#include 
#include 

using namespace std;

const int MAX_N = 500005;

struct node{
	int v, w, next;
}E[MAX_N << 1];
int head[MAX_N], top = 0;

inline void add(int u, int v)
{
	E[++ top].v = v; E[top].w = 1; E[top].next = head[u]; head[u] = top;
}

int N, M, u, v, w;
int go[MAX_N][20], deep[MAX_N], sum[MAX_N];

void dfs(int last, int x)
{
	for (int i=head[x]; i; i=E[i].next){
		if (E[i].v != last){
			go[E[i].v][0] = x; deep[E[i].v] = deep[x] + 1;
			sum[E[i].v] = E[i].w + sum[x];
			dfs(x, E[i].v);
		}
	}
}

void pre(void)
{
	for (int k=1; k<=19; k++)
	  for (int i=1; i<=N; i++)
	  	go[i][k] = go[go[i][k - 1]][k - 1];
}

void goup(int &x, int k)
{
	for (int i=19; i>=0; i--)
		if ((1< deep[v]) goup(u, deep[u] - deep[v]);
	if (deep[v] > deep[u]) goup(v, deep[v] - deep[u]);
	if (u == v) return u;
	for (int k=19; k>=0; k--)
	  if (go[u][k] != go[v][k])
	  	u = go[u][k], v = go[v][k];
	return go[u][0];
}

void doit(void)
{
	while (M --){
		scanf("%d%d%d", &u, &v, &w);
		int f1 = LCA(u, v), f2 = LCA(v, w), f3 = LCA(w, u);
		
		int r, ans = 0, another, a, b;
		if(deep[f1] < deep[f2]){
			r = f2, another = u; a = v, b = w;
		} else { r = f1, another = w; a = u, b = v;}
		if(deep[r] < deep[f3]){
			r = f3, another = v; a = u, b = w;
		}
		
		int fr = LCA(r, another);		
		printf("%d %d\n", r, sum[a] + sum[b] - 2 * sum[r] + sum[r] + sum[another] - 2 * sum[fr]);
	}
}

int main(void)
{
	scanf("%d%d", &N, &M);
	for (int i=1; i<=N-1; i++)
	  scanf("%d%d", &u, &v),
	  add(u, v), add(v, u);
	
	dfs(0, 1); pre();
	doit();
	return 0;
}