D. Arboris Contractio time limit per test2 seconds...
Criado em: 10 de agosto de 2025
Respondido usando GPT-5 Thinking por Chat01
Pergunta
D. Arboris Contractio
time limit per test2 seconds
memory limit per test256 megabytes
Kagari is preparing to archive a tree, and she knows the cost of doing so will depend on its diameter^{\text{∗}}. To keep the expense down, her goal is to shrink the diameter as much as possible first. She can perform the following operation on the tree:
Choose two vertices s and t. Let the sequence of vertices on the simple path^{\text{†}} from s to t be v_0, v_1, \dots, v_k, where v_0 = s and v_k = t.
Remove all edges along the path. In other words, remove edges (v_0, v_1), (v_1, v_2), \dots, (v_{k-1}, v_k).
Connect vertices v_1, v_2, \dots, v_k directly to v_0. In other words, add edges (v_0, v_1), (v_0, v_2), \dots, (v_0, v_k).
It can be shown that the graph is still a tree after the operation.
Help her determine the minimum number of operations required to achieve the minimal diameter.
^{\text{∗}}The diameter of a tree is the longest possible distance between any pair of vertices. The distance itself is measured by the number of edges on the unique simple path connecting them.
^{\text{†}}A simple path is a path between two vertices in a tree that does not visit any vertex more than once. It can be shown that the simple path between any two vertices is always unique.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 \le t \le 10^4). The description of the test cases follows.
The first line of each test case contains one integer n (2 \le n \le 2 \cdot 10^5) — the number of the vertices in the tree.
The following n-1 lines of each test case describe the tree. Each of the lines contains two integers u and v (1 \le u, v \le n, u \neq v) that indicate an edge between vertex u and v. It is guaranteed that these edges form a tree.
It is guaranteed that the sum of n over all test cases does not exceed 2 \cdot 10^5.
Output
For each test case, output one integer — the minimum number of operations to minimize the diameter.
Example
InputCopy
4
4
1 2
1 3
2 4
2
2 1
4
1 2
2 3
2 4
11
1 2
1 3
2 4
3 5
3 8
5 6
5 7
7 9
7 10
5 11
OutputCopy
1
0
0
4
Note
In the first test case, the diameter of the original tree is 3. Kagari can perform an operation on s = 3 and t = 4. As the figure depicts, the operations includes the following steps:
Remove edges (3, 1), (1, 2) and (2, 4).
Add edges (3, 1), (3, 2) and (3, 4).
After the operation, the diameter reduces to 2. It can be shown that 2 is the minimum diameter.
In the second test case, the diameter of the tree is 1. It can be shown that 1 is already the minimum, so Kagari can perform no operation.
give solution in cpp. there should be no comments in the code
Pensamento
Resposta
cpp#include <bits/stdc++.h> using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int t; if(!(cin >> t)) return 0; while(t--) { int n; cin >> n; vector<vector<int>> g(n+1); vector<int> deg(n+1,0); for(int i=0;i<n-1;i++){ int u,v; cin >> u >> v; g[u].push_back(v); g[v].push_back(u); deg[u]++; deg[v]++; } int L = 0; for(int i=1;i<=n;i++) if(deg[i]==1) L++; vector<int> leafNei(n+1,0); for(int i=1;i<=n;i++){ int cnt = 0; for(int j: g[i]) if(deg[j]==1) cnt++; leafNei[i] = cnt; } int ans = n; for(int i=1;i<=n;i++){ int ops = L - leafNei[i] - (deg[i]==1 ? 1 : 0); ans = min(ans, ops); } cout << ans << '\n'; } return 0; }