\(\text{Solution}\)

问题的转化，建成圆方树后，变为询问 \(S\) 在圆方树上对应的连通子图中的圆点个数减去 \(|S|\)
而根据 \(\text{SDOI2015 寻宝游戏}\) 里的一个重要结论
包含 \(S\) 的极小连通子图边权和的两倍等于将 \(S\) 里的点按 \(dfs\) 序排序后 \(dis(a_1,a_2)+dis(a_2,a_3)+...+dis(a_{n-1},a_n)+dis(a_n,a_1)\)
而在圆方树中统计圆点数量，可将圆点到其上方点的边权赋为 \(1\)
然后按结论计算后结果除以 \(2\) 相当于一棵树的若干边被计算了，一条边对应其下一个圆点
而如果最浅的 \(LCA\) 是圆点，我们还要统计它
注意多组数据，倍增求 \(LCA\) 的 \(fa\) 数组要清零！！

\(\text{Code}\)

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
#define IN inline
#define RE register
using namespace std;

const int N = 2e5 + 5;
int T, n, m, q, cnt, a[N], h1[N], h2[N], tot1, tot2, low[N], dfn[N], dfc, top, stk[N], fa[N][21], dis[N], dep[N];
struct edge{int to, nxt;}e1[N * 4], e2[N * 4];
IN void add1(int x, int y){e1[++tot1] = edge{y, h1[x]}, h1[x] = tot1;}
IN void add2(int x, int y){e2[++tot2] = edge{y, h2[x]}, h2[x] = tot2;}
IN bool cmp(int x, int y){return dfn[x] < dfn[y];}

void Tarjan(int x)
{
	dfn[x] = low[x] = ++dfc, stk[++top] = x;
	for(RE int i = h1[x]; i; i = e1[i].nxt)
	{
		int v = e1[i].to;
		if (!dfn[v])
		{
			Tarjan(v), low[x] = min(low[x], low[v]);
			if (dfn[x] == low[v])
			{
				++cnt, add2(cnt, x), add2(x, cnt);
				for(RE int u = 0; u ^ v; --top) u = stk[top], add2(cnt, u), add2(u, cnt);
			}
		}
		else low[x] = min(low[x], dfn[v]);
	}
}

void Dfs(int x)
{
	for(RE int i = 1; i < 19; i++) if (fa[x][i - 1]) fa[x][i] = fa[fa[x][i - 1]][i - 1]; else break;
	dis[x] += (x <= n), dfn[x] = ++dfc;
	for(RE int i = h2[x]; i; i = e2[i].nxt)
	{
		int v = e2[i].to;
		if (v == fa[x][0]) continue;
		fa[v][0] = x, dis[v] += dis[x], dep[v] = dep[x] + 1, Dfs(v);
	}
}

IN int LCA(int x, int y)
{
	if (dep[x] < dep[y]) swap(x, y);
	int deep = dep[x] - dep[y];
	for(RE int i = 0; i < 19; i++) if ((deep >> i) & 1) x = fa[x][i];
	if (x == y) return x;
	for(RE int i = 18; i >= 0; i--) if (fa[x][i] ^ fa[y][i]) x = fa[x][i], y = fa[y][i];
	return fa[x][0];
}
IN int Query(int x, int y){return dis[x] + dis[y] - dis[LCA(x, y)] * 2;}

IN void read(int &x)
{
	x = 0; int f = 1; char ch = getchar();
	for(; !isdigit(ch); f = (ch == '-' ? -1 : f), ch = getchar());
	for(; isdigit(ch); x = (x<<3)+(x<<1)+(ch^48), ch = getchar());
	x *= f;
}

int main()
{
	read(T);
	for(; T; --T)
	{
		read(n), read(m), cnt = n, top = dfc = tot1 = tot2 = 0;
		memset(h1, 0, sizeof h1), memset(h2, 0, sizeof h2), memset(dis, 0, sizeof dis);
		memset(fa, 0, sizeof fa), memset(dfn, 0, sizeof dfn);
		for(RE int i = 1, u, v; i <= m; i++) read(u), read(v), add1(u, v), add1(v, u);
		Tarjan(1), dfc = 0, Dfs(1), read(q);
		for(RE int num; q; --q)
		{
			read(num); for(RE int i = 1; i <= num; i++) read(a[i]);
			sort(a + 1, a + num + 1, cmp); int ans = -2 * num;
			for(RE int i = 1; i <= num; i++) ans += Query(a[i], a[i % num + 1]);
			if (LCA(a[1], a[num]) <= n) ans += 2;
			printf("%d\n", ans >> 1);
		}
	}
}

标签：cnt,游戏,int,战略,fa,dfn,low,SDOI2018,dis
来源： https://www.cnblogs.com/leiyuanze/p/15844695.html