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>
using namespace std;
using i64 = long long;
#define L(i,j,k) for(int i=(j);i<=(k);++i)
#define R(i,j,k) for(int i=(j);i>=(k);--i)
#define rep(i, n) L(i, 1, n)
#define all(x) x.begin(),x.end()
#define me(x,a) memset(x,a,sizeof(x))
#include<random>
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int nax = 1E5 + 5;
int N, a[nax], dp[nax][2][2];
vector<int> G[nax];
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);
cin >> N;
L(i, 1, N - 1) {
int u, v; cin >> u >> v;
G[u].push_back(v);
G[v].push_back(u);
}
L(i, 1, N) cin >> a[i];
function<void(int, int)> dfs = [&](int u, int f) {
for (auto & v : G[u]) {
if (v == f) continue;
dfs(v, u);
}
/* each node is only pressed once and we need to pass on info to the parent node
* dp[u][A][B] = minimum amount of operations done when we have considered node u
* and we have A = whether we have pressed node u and B = have we pressed an odd or even number of buttons
*/
for (int press = 0; press < 2; press++) { // do I want to press this button
for (int par = 0; par < 2; par++) { // whether we want to pass on to the parent
int even = 0, odd = N * 2; // initially we only pressed even number of buttons
for (auto & v : G[u]) {
if (v == f) continue;
// dp[v][whether we pressed child][passing to parent of v that is u (our state whether we want to press)]
int _od = min(odd + dp[v][0][press], even + dp[v][1][press]); // either odd + 1 -> even or even + 1 -> odd
int _ev = min(odd + dp[v][1][press], even + dp[v][0][press]); // opposite way
swap(_od, odd); swap(_ev, even);
}
if (press ^ par ^ a[u]) dp[u][press][par] = odd + press; // press ^ par ^ a[u] == 1, we need to perform an operation
else dp[u][press][par] = even + press; // otherwise
// the value is stored in odd when we want to change because pressing a odd amount of buttons will cause the button to be flipped
}
}
};
dfs(1, 1);
int ans = min(dp[1][1][0], dp[1][0][0]); // since the third one shows how the parent would be pressed, there is no parent so no need to consider
if (ans > N) cout << "impossible\n";
else cout << ans << '\n';
}
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