NC14250 MMSet2

题目链接

题意：
$给一棵n个点的树，点的编号为1，2，3，\dots \dots ，n$
$q次询问，每次询问给定一个点集S$
$令f(u)为u到点集内点的最大距离$
$求mi{n}_{u=1..n}f(u)$
题解：
$n<=3e5,\sum S<=1e6$
$题意转化：$
$找到一个点到点集点的最大距离最小$
$那么肯定要找点集最远的两个点$
$然后找一个点到他俩距离相同或者差为1$
$所以只要将点集的直径平分即可$
$那么最后答案就是ceil(直径/2)$
$可以考虑用虚树建树找直径$
$但是可以通过求直径的性质$
$直径的一个端点肯定是深度最深的点$
$找到这个点，然后利用LCA求出到其他点的最大距离，除二即可$
AC代码

/*
    Author : zzugzx
    Lang : C++
    Blog : blog.csdn.net/qq_43756519
*/
#include
using namespace std;

#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define endl '\n'
#define SZ(x) (int)x.size()
#define mem(a, b) memset(a, b, sizeof(a))

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int mod = 1e9 + 7;
//const int mod = 998244353;

const double eps = 1e-6;
const double pi = acos(-1.0);
const int maxn = 1e6 + 10;
const int N = 1e2 + 5;
const ll inf = 0x3f3f3f3f;
const ll oo = 8e18;
const int dir[][2]={
     {
     0, 1}, {
     1, 0}, {
     0, -1}, {
     -1, 0}, {
     1, 1}, {
     1, -1}, {
     -1, 1}, {
     -1, -1}};
int n,depth[maxn], f[maxn][50];
int from[maxn], to[maxn << 1], nxt[maxn << 1], cnt,Log[maxn];
//链式前向星加边
void addEdge (int u, int v) {
     
    to[++cnt] = v, nxt[cnt] = from[u], from[u] = cnt;
}
//计算深度&计算祖先
void dfs (int u, int fa) {
     
    depth[u] = depth[fa] + 1;
    for (register int i = 1; i <= Log[n]; ++i) {
     
        if ((1 << i) > depth[u]) break;
        f[u][i] =  f[f[u][i - 1]][i - 1];
    }
    for (register int i = from[u]; i; i = nxt[i]) {
     
        ll v = to[i];
        if (v == fa) continue;
        f[v][0] = u;
        dfs (v, u);
    }
}
//计算LCA
inline int LCA (int x, int y) {
     
    if (depth[x] < depth[y]) swap(x, y);
    //我们默认x为更深的那个点
    for(register int i = Log[n] ; i >= 0 ; --i)
		if(depth[x] - (1 << i) >= depth[y]) x = f[x][i];
    //将x跳到和y同一深度上
    if (x == y) return x;
    for (register int i = Log[n]; i >= 0; --i)
        if (f[x][i] != f[y][i])
            x = f[x][i], y = f[y][i];
    //一起向上跳
    return f[x][0];
    //不难看出，此时两个点均在其LCA的下方，往上跳一次即可
}
void init(){
     
	Log[0] = -1;
	cnt = 0;
	for(int i = 1; i <= n; i++){
     
        from[i] = 0,depth[i] = 0;
        for(int j = 0; j < 50; j++)
            f[i][j] = 0;
	}
    for (register int i = 1, u, v; i < n; ++i) {
     
        cin >> u >> v;
        addEdge (u, v); addEdge(v, u);
        Log[i] = Log[i >> 1] + 1;
    }
    Log[n] = Log[n >> 1] + 1;
    dfs(1,0);
}
int dis(int p , int q){
     return depth[p] + depth[q] - 2 * depth[LCA(p , q)];}
int main()
{
     
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
//  freopen("in.txt", "r", stdin);
//  freopen("out.txt", "w", stdout);
    cin >> n;
    init();
    int _;
    cin >> _;
    while (_--) {
     
        int k, x, y = 0, ans = 0;
        cin >> k;
        vector<int> v;
        for(int i = 1; i <= k; i++) {
     
            cin >> x;
            if (depth[x] > depth[y]) y = x;
            v.pb(x);
        }
        for (auto i : v) ans = max(ans, dis(y, i));
        cout << (ans + 1) /2 << endl;
    }
    return 0;
}