P4320 道路相遇

考虑建出圆方树。

对于任意两个点，其路径中必须要经过的点就是这两个点对应的圆方树中点的路径之间圆点数量。也就是割点数量。

于是 LCA 处理即可。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <vector>
using namespace std;

const int N = 1e6 + 5;

int dfn[N], low[N], idx, cnt;
int n, m, q;
vector<int> G[N], NG[N];
int sum[N];
int stk[N], top;

void tarjan(int u)
{
	dfn[u] = low[u] = ++idx;
	stk[++top] = u;
	for (int j : G[u])
	{
		if (!dfn[j])
		{
			tarjan(j);
			low[u] = min(low[u], low[j]);
			if (dfn[u] == low[j])
			{
				cnt++;
				for (int y = 0; y != j; top--)
				{
					y = stk[top];
					NG[y].emplace_back(cnt);
					NG[cnt].emplace_back(y);
				}
				NG[cnt].emplace_back(u);
				NG[u].emplace_back(cnt);
			}
		}
		else
		{
			low[u] = min(low[u], dfn[j]);
		}
	}
}

void dfs(int u, int fa)
{
	sum[u] = sum[fa] + (u <= n);
	for (int j : NG[u])
	{
		if (j != fa)
		{
			dfs(j, u);
		}
	}
}

class TreeCut
{
public:
	int id[N], top[N], sz[N], dep[N], fa[N], son[N], idx;
	void dfs1(int u, int f)
	{
		dep[u] = dep[f] + 1;
		fa[u] = f;
		sz[u] = 1;
		for (int j : NG[u])
		{
			if (j != f)
			{
				dfs1(j, u);
				sz[u] += sz[j];
				if (sz[son[u]] < sz[j]) son[u] = j;
			}
		}
	}
	void dfs2(int u, int tf)
	{
		id[u] = ++idx;
		top[u] = tf;
		if (!son[u]) return;
		dfs2(son[u], tf);
		for (int j : NG[u])
		{
			if (j != son[u] && j != fa[u])
			{
				dfs2(j, j);
			}
		}
	}
	void build()
	{
		dfs1(1, 1);
		dfs2(1, 1);
	}
	int LCA(int u, int v)
	{
		while (top[u] ^ top[v])
		{
			if (dep[top[u]] < dep[top[v]]) swap(u, v);
			u = fa[top[u]];
		}
		return (dep[u] < dep[v] ? u : v);
	}
}tc;

int main()
{
	scanf("%d%d", &n, &m);
	cnt = n;
	for (int i = 1; i <= m; i++)
	{
		int u, v;
		scanf("%d%d", &u, &v);
		G[u].emplace_back(v);
		G[v].emplace_back(u);
	}
	tarjan(1);
	dfs(1, 1);
	tc.build();
	scanf("%d", &q);
	while (q--)
	{
		int s, t;
		scanf("%d%d", &s, &t);
		int lca = tc.LCA(s, t);
		printf("%d\n", sum[s] + sum[t] - 2 * sum[lca] + (lca <= n));
	}
	return 0;
}