LCA (模板，HDOJ 2586 How far away? ）

多次查询树上两点之间的距离。

先预处理bfs，求出树根到各个点的距离和各个点的深度，同时更新f[maxn][20],f(x,k)表示x的2^k辈祖先

f(x,0)就是x的父节点。对于任意k,[1,logn], f[x,k]=f[f[x,k-1],k-1]。

求lca时，将x,y调整至同一深度，再逐步向上寻找。

#include <iostream>
#include <map>
#include <vector>
#include <cstring>
#include <cstdio>
#include <queue>
#include <algorithm>
#define LL long long
using namespace std;
const int maxn = 100005;
int n, m, t, cnt;
int f[maxn][20];
int head[maxn];
LL d[maxn], dis[maxn];
struct node
{
	int u, v, w, nex;
}edge[maxn];
void addedge(int u, int v, int w)
{
	edge[cnt].u = u;
	edge[cnt].v = v;
	edge[cnt].w = w;
	edge[cnt].nex = head[u];
	head[u] = cnt++;
}
void init()
{
	memset(f, 0, sizeof(f));
	for (int i = 0; i < maxn; i++)
	{
		head[i]=-1;
		d[i] = 0;
	}
	cnt = 0;
}
void bfs()
{
	queue<int>q;
	q.push(1);
	d[1] = 1;
	while (!q.empty())
	{
		int h = q.front();
		q.pop();
		for (int i = head[h]; i != -1; i = edge[i].nex)
		{
			int v = edge[i].v;
			int w = edge[i].w;
			if (d[v]) continue;
			d[v] = d[h] + 1;
			dis[v] = dis[h] + w;
			f[v][0] = h;
			for (int j = 1; j <= t; j++)
				f[v][j] = f[f[v][j - 1]][j - 1];
			q.push(v);
		}
	}
}
int lca(int x, int y)
{
	if (d[x] > d[y]) swap(x, y);
	for (int i = t; i >= 0; i--)
		if (d[f[y][i]] >= d[x])
			y = f[y][i];
	if (x == y)
		return x;
	for (int i = t; i >= 0; i--)
		if (f[x][i] != f[y][i])
			x = f[x][i], y = f[y][i];
	return f[x][0];
}
int main()
{
	int T;
	ios::sync_with_stdio(false);
	cin >> T;
	int cnt = 1;
	while (T--)
	{
		cin >> n >> m;
		t = (int)(log(n) / log(2)) + 1;//树的可能最大深度
		init();
		for (int i = 0; i < n - 1; i++)
		{
			int a, b, c;
			cin >> a >> b >> c;
			addedge(a, b, c);
			addedge(b, a, c);
		}
		bfs();
		for (int i = 0; i < m; i++)
		{
			int x, y;
			cin >> x >> y;
			LL ans = dis[x] + dis[y] - 2 * dis[lca(x, y)];
			cout << ans << endl;
		}
		//cout << endl;
	}
	return 0;
}