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;
template<typename T>
void out(T x) { cout << x << endl; exit(0); }
#define watch(x) cout << (#x) << " is " << (x) << endl
using ll = long long;
const int maxn = 1e6 + 5;
int n;
ll a[maxn];
vector<int> g[maxn];
int main() {
ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
cin>>n;
for (int i=1; i<=n; i++) {
cin>>a[i];
}
ll res = 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);
res += max(a[u],a[v]);
}
ll sum = 0;
ll hi = 0;
for (int i=1; i<=n; i++) {
hi = max(hi,a[i]);
sum += a[i];
}
res += sum;
res -= hi;
out(res);
return 0;
}
// [1] it's optimal to always snip off an edge from the max node in the tree (swapping argument)
// sum over edges e { max(e.u, e.v) }
// [2] every node is eventually isolated and snipped off
// sum a[i]
// subtract max a[i] because it's contribution is entirely from [1]
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