MMSet2

思路

这道题目显然能够通过$3\times 10^5\times 10^6$的复杂度来暴力，这显然不能达到题目要求的复杂度，因此我们可以对题目要求我们计算的东西进行转换。

某个点到所有点集的最大距离最小，这就有点像是重心的求法了，但是这题又有所不同，如果这是在一颗树上，显然我们可以很快的得到答案，$ans=\lceil \frac{直径}{2}\rceil$，所以这题我们也可以转换思想，每次求解得到点集中两点之间最长的距离，然后再对他向上取整。

问题转换为求解点集中得直径了，所以我们可以找到点集中的深度最大的点，然后通过这个点去求得点集的直径。

代码(事实证明树剖求lca是真的快)

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int N = 3e5 + 10;int head[N], to[N << 1], nex[N << 1], cnt = 1;int dep[N], son[N], sz[N], fa[N], top[N], tot;void dfs1(int rt, int f) {dep[rt] = dep[f] + 1;sz[rt] = 1, fa[rt] = f;for(int i = head[rt]; i; i = nex[i]) {if(to[i] == f)  continue;dfs1(to[i], rt);if(!son[rt] || sz[to[i]] > sz[son[rt]])son[rt] = to[i];sz[rt] += sz[to[i]];}
}void dfs2(int rt, int t) {top[rt] = t;if(!son[rt])    return ;dfs2(son[rt], t);for(int i = head[rt]; i; i = nex[i]) {if(to[i] == fa[rt] || to[i] == son[rt]) continue;dfs2(to[i], to[i]);}
}int lca(int x, int y) {while(top[x] != top[y]) {if(dep[top[x]] < dep[top[y]]) swap(x, y);x = fa[top[x]];}return dep[x] < dep[y] ? x : y;
}int dis(int x, int y) {return dep[x] + dep[y] - 2 * dep[lca(x, y)];
}void add(int x, int y) {to[cnt] = y;nex[cnt] = head[x];head[x] = cnt++;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int n = read();for(int i = 1; i < n; i++) {int x = read(), y = read();add(x, y);add(y, x);}dfs1(1, 0);dfs2(1, 1);int Q = read();for(int i = 1; i <= Q; i++) {int m = read();vector<int> a;int u = 0;for(int j = 1; j <= m; j++) {int x = read();if(dep[x] > dep[u]) u = x;a.pb(x);}int ans = 0;for(int v : a) {if(v == u) continue;ans = max(ans, dis(u, v));}printf("%d\n", (ans + 1) >> 1);}return 0;
}