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;
const int lim = 3e5 + 5;
bool on[lim], vis[lim];
vector<int> edges[lim];
long long dp[lim][2][2];
void dfs(int nd) {
vis[nd] = 1;
if(on[nd])
dp[nd][1][0] = 1, dp[nd][0][1] = 0;
else
dp[nd][0][0] = 0, dp[nd][1][1] = 1;
for(auto i : edges[nd]) {
if(!vis[i]) {
dfs(i);
// case now not pressed
// awalannya 0
int x = dp[nd][0][0], y = dp[nd][0][1];
dp[nd][0][0] = min(dp[i][0][0] + x, dp[i][1][0] + y);
dp[nd][0][1] = min(dp[i][0][0] + y, dp[i][1][0] + x);
// case now pressed
// awalannya 1
x = dp[nd][1][0], y = dp[nd][1][1];
dp[nd][1][0] = min(dp[i][0][1] + x, dp[i][1][1] + y);
dp[nd][1][1] = min(dp[i][0][1] + y, dp[i][1][1] + x);
}
}
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(0);
int n;
cin >> n;
for(int i = 0; i < n - 1; ++i) {
int u, v;
cin >> u >> v;
edges[u].push_back(v);
edges[v].push_back(u);
}
for(int i = 1; i <= n; ++i)
cin >> on[i];
for(int i = 1; i <= n; ++i) {
for(int j = 0; j < 2; ++j)
dp[i][j][0] = dp[i][j][1] = 1e6;
}
// dp on tree
// cari dr child
// pake dp[i][1], nandain dp di subtree ini dengan root 1 atau root 0
// problem -> child bisa effect root
// nanti ans itu dp[1][0]
// nanti 4 kemungkinan
// root 0, root changed
// root 0, root not changed
// root 1, root changed
// root 1, root not changed
// transisi dp?
dfs(1);
if(min(dp[1][1][0], dp[1][0][0]) > n) {
cout << "impossible" << endl;
}
else
cout << min(dp[1][1][0], dp[1][0][0]) << endl;
}
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