CF1695D2. Tree Queries (Hard Version) (树形DP, 思维)

https://codeforces.com/contest/1695/problem/D2
题意：

思路：

• 找任意度数大于3的点u做根，u有c个子树，则c-1个子树都需要有询问点
• 这是因为，度数大于3的点至少有两棵子树有询问点，这样对于一个子树，子树外有询问点，只需关注子树内

树上问题以一个子树为一个阶段进行分析

#include<bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(false) ,cin.tie(0), cout.tie(0);
//#pragma GCC optimize(3,"Ofast","inline")
#define ll long long
#define PII pair<int, char>
//#define int long long
const int N = 2e5 + 5;
const int M = 1e5 + 5;
const int INF = 0x3f3f3f3f;
const ll LNF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double PI = acos(-1.0);
vector<int> E[N];
bool f[N];
int dfs ( int u, int fa ) {
 f[u] = 0; int res = 0; int cnt = 0;
 for ( auto v : E[u] ) {
   if( v == fa ) continue;
   res += dfs(v, u);
   f[u] |= f[v];
   cnt += !f[v];
 }
 if( cnt >= 2 ) res += cnt - 1, f[u] = 1;
 return res;
}
void solve() {
 int n; cin >> n;
 if( n == 1 ) {
   cout << 0 << '\n'; return;
 }
 for ( int i = 1; i <= n; ++ i ) E[i].clear();
 for ( int i = 1; i <= n - 1; ++ i ) {
   int u,v; cin >> u >> v;
   E[u].push_back(v), E[v].push_back(u);
 }
 int root = 0;
 for ( int i = 1; i <= n; ++ i ) {
   if(E[i].size() >= 3) {
     root = i; break;
   }
 }
 if( !root ) {
   cout << 1 << '\n'; return;
 }
 cout << dfs(root, -1) << '\n';

}
int main () {
 IOS
 int t = 1; cin >> t;
 while ( t -- ) solve();
 return 0;
}