HDU 2874 LCA

#include 
#include 
#include 
#include 
using namespace std;
const int maxn = 1E5 + 10;
const int maxm = maxn * 2;
const int maxq = 2E6 + 10;
int n, m, q, dis[maxn], vis[maxn], father[maxn], ans[maxq / 2], a, b, c;
struct Edge
{
	int v, len, next;
};
Edge edge[maxm];
int tot, head[maxn];
void addedge(int a, int b, int c)
{
	edge[tot].v = b; edge[tot].len = c;
	edge[tot].next = head[a]; head[a] = tot++;
}
struct Ques
{
	int v, index, next;
};
Ques Q[maxq];
int q_tot, q_head[maxn];
void addques(int a, int b, int index)
{
	Q[q_tot].v = b; Q[q_tot].index = index;
	Q[q_tot].next = q_head[a]; q_head[a] = q_tot++;
}
int Find(int k)
{
	if (father[k] == k) return k;
	else return father[k] = Find(father[k]);
}
void Tarjan_LCA(int k, int deep, int root)
{
	father[k] = k;
	vis[k] = root;
	dis[k] = deep;
	for (int j = head[k]; j != -1; j = edge[j].next)
	{
		int v = edge[j].v;
		if (vis[v] == -1)
		{
			Tarjan_LCA(v, deep + edge[j].len, root);
			father[v] = k;
		}
	}
	for (int j = q_head[k]; j != -1; j = Q[j].next)
	{
		int v = Q[j].v;
		if (vis[v] == root)
			ans[Q[j].index] = dis[v] + dis[k] - 2 * dis[Find(v)];
	}
}
int main(int argc, char const *argv[])
{
	while (~scanf("%d%d%d", &n, &m, &q))
	{
		tot = q_tot = 0;
		memset(head, -1, sizeof(head));
		memset(q_head, -1, sizeof(q_head));
		memset(vis, -1, sizeof(vis));
		for (int i = 0; i < m; i++)
		{
			scanf("%d%d%d", &a, &b, &c);
			addedge(a, b, c);
			addedge(b, a, c);
		}
		for (int i = 0; i < q; i++)
		{
			scanf("%d%d", &a, &b);
			ans[i] = -1;
			addques(a, b, i);
			addques(b, a, i);
		}
		for (int i = 1; i <= n; i++) if (vis[i] == -1) Tarjan_LCA(i, 0, i);
		for (int i = 0; i < q; i++)
			if (ans[i] == -1) printf("Not connected\n");
			else printf("%d\n", ans[i]);
	}
	return 0;
}

询问 a，b 之间的最短距离。因为图中不可能出现环，所以两点之间有且仅有一条路，或者没有。

预处理出ab的祖先和距离，直接返回dis[a]+dis[b]-2*dis[father];