CodeForces 1946C Tree Cutting
Tree Cutting 考虑从叶子节点开始切割，每次切割的时候只能切掉整颗子树，因此从叶子节点统计当前子树的节点的个数，当子树大小至少为 $x$ 的时候进行一次切割，最后切掉的子树的数量不少于 $k$。因为是从叶子节点考虑，所以可以考虑拓扑排序的过程，用 BFS 的思想从叶子开始切割。 更好的实现方式是从子树节点个数的角度考虑，这是一个经典的 DFS 问题，当子树的大小至少为 $x$ 时，此时进行一次切割，这个子树的大小不贡献给父节点。 DFS 实现，DFS 的过程中只需要记录上一次访问的节点就可以保证不重复访问。 // Date: Sat Mar 23 09:39:02 2024 #include <climits> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <functional> #include <iomanip> #include <iostream> #include <map> #include <queue> #include <set> #include <sstream> #include <stack> #include <string> #include <utility> #include <vector> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef vector<int> VI; typedef pair<int, int> PII; typedef pair<ll, ll> PLL; template <class T> using pq = priority_queue<T>; template <class T> using pqg = priority_queue<T, vector<T>, greater<T>>; const int INF = 0x3f3f3f3f, MOD = 1e9 + 7, MOD1 = 998'244'353; const ll INFL = 0x3f3f3f3f'3f3f3f3f; const double eps = 1e-8; const int dir[8][2] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}, }; const ull Pr = 131; #define For(i, a, b) for (int i = int(a); i < int(b); ++i) #define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i) #define For1(i, a, b) for (int i = int(a); i <= int(b); ++i) #define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i) #define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i]) #define f1 first #define f2 second #define pb push_back #define has(a, x) (a.find(x) != a.end()) #define nemp(a) (!a.empty()) #define all(a) (a).begin(), (a).end() #define SZ(a) int((a).size()) #define NL cout << '\n'; template <class T> bool ckmin(T &a, const T &b) { return b < a ? a = b, 1 : 0; } template <class T> bool ckmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; } template <typename t> istream &operator>>(istream &in, vector<t> &vec) { for (t &x : vec) in >> x; return in; } template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) { int n = SZ(vec); For(i, 0, n) { out << vec[i]; if (i < n - 1) out << ' '; } return out; } // int128 input and output #ifdef _DEBUG using lll = __int128; istream &operator>>(istream &is, lll &v) { string s; is >> s; v = 0; for (auto &it : s) if (isdigit(it)) v = v * 10 + it - '0'; if (s[0] == '-') v *= -1; return is; } ostream &operator<<(ostream &os, const lll &v) { if (v == 0) return (os << "0"); lll num = v; if (v < 0) os << '-', num = -num; string s; for (; num > 0; num /= 10) s.pb((char)(num % 10) + '0'); reverse(all(s)); return (os << s); } #endif // end of int128 void __print(int x) { cerr << x; } void __print(long x) { cerr << x; } void __print(long long x) { cerr << x; } void __print(unsigned x) { cerr << x; } void __print(unsigned long x) { cerr << x; } void __print(unsigned long long x) { cerr << x; } void __print(float x) { cerr << x; } void __print(double x) { cerr << x; } void __print(long double x) { cerr << x; } void __print(char x) { cerr << '\'' << x << '\''; } void __print(const char *x) { cerr << '\"' << x << '\"'; } void __print(const string &x) { cerr << '\"' << x << '\"'; } void __print(bool x) { cerr << (x ? "true" : "false"); } template <typename T, typename V> void __print(const pair<T, V> &x) { cerr << '{'; __print(x.first); cerr << ", "; __print(x.second); cerr << '}'; } template <typename T> void __print(const T &x) { int f = 0; cerr << '{'; for (auto &i : x) cerr << (f++ ? ", " : ""), __print(i); cerr << "}"; } void _print() { cerr << "]\n"; } template <typename T, typename... V> void _print(T t, V... v) { __print(t); if (sizeof...(v)) cerr << ", "; _print(v...); } #ifdef _DEBUG #define debug1(x) cout << #x " = " << x << endl; #define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl; #define debug3(x, y, z) \ cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl; #define dbg(x...) \ cerr << "\e[91m" << __func__ << ":" << __LINE__ << " [" << #x << "] = ["; \ _print(x); \ cerr << "\e[39m" << endl; #else #define debug1 #define debug2 #define debug3 #define dbg(x...) #endif const int N = 100100, M = 2 * N; int n, k, u, v, h[N], cnt; int idx, e[M], ne[M]; void Init() { idx = 0; memset(h, -1, sizeof h); } void Add(int a, int b) { e[idx] = b, ne[idx] = h[a], h[a] = idx++; } int dfs(int u, int pre, int x) { int sum = 1; ForE(i, u) { int j = e[i]; if (j == pre) continue; sum += dfs(j, u, x); } if (sum >= x) { cnt++; sum = 0; } return sum; } bool check(int x) { cnt = 0; dfs(1, -1, x); return cnt >= k + 1; } void solve() { cin >> n >> k; Init(); For(i, 1, n) { cin >> u >> v; Add(u, v); Add(v, u); } int l = 1, r = 1e5 + 10, mid; while (l < r) { mid = (l + r + 1) / 2; if (check(mid)) l = mid; else r = mid - 1; } cout << l << '\n'; } int main(void) { #ifdef _DEBUG freopen("1946c.in", "r", stdin); #endif std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int T = 1; cin >> T; while (T--) { solve(); } return 0; } BFS 实现，类似拓扑排序的思想，把度数为 $1$ 的叶子节点加入队列，此时需要记录节点是否已经被访问过。 ...