题意:
题解:
AC代码
/* Author : zzugzx Lang : C++ Blog : blog.csdn.net/qq_43756519 */ #include<bits/stdc++.h> 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; }