이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "roads.h"
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 5;
multiset<long long> st[MAXN];
long long sum[MAXN], dp[MAXN][2];
int deg[MAXN], vis[MAXN];
vector<pair<int, int>> adj[MAXN];
inline void upd(int u, long long v, int op) {
sum[u] += op * v;
if (op == +1)
st[u].insert(v);
if (op == -1)
st[u].erase(st[u].find(v));
}
void solve(int u) {
vector<long long> add, del;
long long tot = 0;
int X = vis[u], lim = deg[u] - X; // need to del
while ((int)st[u].size() > max(lim, 0))
upd(u, *st[u].rbegin(), -1);
for (auto [v, w] : adj[u]) {
if (deg[v] <= X) break;
if (vis[v] == X) continue;
vis[v] = X;
solve(v);
tot += dp[v][0];
long long val = dp[v][1] + w - dp[v][0];
if (val > 0) {
add.push_back(val), upd(u, val, +1);
} else {
tot += val, lim --;
}
}
for (int i = 0; i < 2; i++) {
while ((int)st[u].size() > max(lim - i, 0))
del.push_back(*st[u].rbegin()), upd(u, *st[u].rbegin(), -1);
dp[u][i] = tot + sum[u];
}
for (auto v : del) upd(u, v, +1);
for (auto v : add) upd(u, v, -1);
}
vector<long long> minimum_closure_costs(int N, vector<int> U, vector<int> V, vector<int> W) {
for (int i = 0; i < N - 1; i++) {
deg[U[i]] ++, deg[V[i]] ++;
adj[U[i]].emplace_back(V[i], W[i]);
adj[V[i]].emplace_back(U[i], W[i]);
}
for (int i = 0; i < N; i++)
sort(adj[i].begin(), adj[i].end(), [&](const pair<int, int> &x, const pair<int, int> &y){ return deg[x.first] > deg[y.first]; });
vector<long long> res(N);
res[0] = accumulate(W.begin(), W.end(), 0ll);
vector<int> p(N);
iota(p.begin(), p.end(), 0);
sort(p.begin(), p.end(), [&](const int &i, const int &j){
return deg[i] < deg[j];
});
for (int X = 1, i = 0; X < N; X++) {
for (; i < N && deg[p[i]] == X; i++) {
for (auto [v, w] : adj[p[i]]) {
if (deg[v] <= X) break;
upd(v, w, 1);
}
}
long long ans = 0;
vector<int> pro;
for(int j = i; j < N; j ++) pro.push_back(j);
random_shuffle(pro.begin(), pro.end());
for(auto j : pro){
if (vis[p[j]] != X) {
vis[p[j]] = X;
solve(p[j]);
ans += dp[p[j]][0];
}
}
for (int j = i; j < N; j++)
res[X] = ans;
}
return res;
}
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