이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
*/
using namespace std;
// using namespace __gnu_pbds;
using namespace chrono;
// mt19937 rng((int) std::chrono::steady_clock::now().time_since_epoch().count());
/*
template <class T> using ordered_set = tree <T, null_type, less <T>, rb_tree_tag, tree_order_statistics_node_update>;
*/
//***************** CONSTANTS *****************
const int MXN = 500'000;
//***************** GLOBAL VARIABLES *****************
vector<int> g[MXN];
map<int, int> mp[MXN];
int deg[MXN], st[MXN], cnt[MXN];
//***************** AUXILIARY STRUCTS *****************
//***************** MAIN BODY *****************
void dfs(int u, int p){
if(cnt[st[u]] != 1) mp[u][st[u]] = 1;
for(int v : g[u]) if(v != p){
dfs(v, u);
if(!mp[v].size()){
deg[u]++;
deg[v]++;
}
else{
deg[u] += deg[v];
deg[v] = 0;
}
if(mp[v].size() > mp[u].size()) swap(mp[u], mp[v]);
for(auto& [state, count] : mp[v]){
mp[u][state] += count;
if(mp[u][state] == cnt[state]) mp[u].erase(state);
}
}
}
void solve(){
int N, K;
cin >> N >> K;
for(int i = 0; i < N - 1; i++){
int u, v;
cin >> u >> v;
--u, --v;
g[u].push_back(v);
g[v].push_back(u);
}
for(int i = 0; i < N; i++){
cin >> st[i];
--st[i];
cnt[st[i]]++;
}
dfs(0, 0);
int ans = 0;
for(int i = 0; i < N; i++) ans += (deg[i] == 1);
cout << (ans + 1) / 2 << '\n';
}
//***************** *****************
int32_t main(){
ios_base::sync_with_stdio(NULL);
cin.tie(NULL);
#ifdef LOCAL
auto begin = high_resolution_clock::now();
#endif
int tc = 1;
// cin >> tc;
for (int t = 0; t < tc; t++)
solve();
#ifdef LOCAL
auto end = high_resolution_clock::now();
cout << fixed << setprecision(4);
cout << "Execution Time: " << duration_cast<duration<double>>(end - begin).count() << "seconds" << endl;
#endif
return 0;
}
/*
If code gives a WA, check for the following :
1. I/O format
2. Are you clearing all global variables in between tests if multitests are a thing
3. Can you definitively prove the logic
*/
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