This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
template <class T>
using Vec = std::vector<T>;
using ll = long long;
template <class F>
struct RecLambda: private F {
explicit RecLambda(F &&f): F(std::forward<F>(f)) { }
template <class... Args>
decltype(auto) operator () (Args&&... args) const {
return F::operator()(*this, std::forward<Args>(args)...);
}
};
struct SlopeTrick {
ll last_slope, last_y;
std::priority_queue<ll> changes;
SlopeTrick(): last_slope(), last_y(), changes() {
changes.push(0);
}
void absorb(SlopeTrick &&other) {
const auto x1 = changes.top();
const auto x2 = other.changes.top();
last_y += other.last_y + (x1 < x2 ? last_slope * (x2 - x1) : other.last_slope * (x1 - x2));
last_slope += other.last_slope;
if (changes.size() < other.changes.size()) {
std::swap(changes, other.changes);
}
while (!other.changes.empty()) {
changes.push(other.changes.top());
other.changes.pop();
}
}
bool pop_last() {
const auto x1 = changes.top();
changes.pop();
if (changes.empty()) return false;
const auto x2 = changes.top();
last_slope -= 1;
last_y += last_slope * (x2 - x1);
return true;
}
};
int main() {
int N, M;
std::cin >> N >> M;
Vec<Vec<std::pair<int, int>>> children(N + M);
for (int i = 1; i < N + M; ++i) {
int p, c;
std::cin >> p >> c;
p -= 1;
children[p].emplace_back(i, c);
}
Vec<SlopeTrick> data(N);
RecLambda([&](auto dfs, const int u) -> void {
for (const auto [v, c]: children[u]) {
if (v >= N) {
SlopeTrick trick;
trick.last_slope = 1;
trick.changes.push(c);
trick.changes.push(c);
data[u].absorb(std::move(trick));
}
else {
dfs(v);
while (data[v].last_slope > 1) {
data[v].pop_last();
}
const auto x1 = data[v].changes.top();
data[v].changes.pop();
const auto x2 = data[v].changes.top();
data[v].changes.pop();
data[v].changes.push(x2 + c);
data[v].changes.push(x1 + c);
data[u].absorb(std::move(data[v]));
}
}
})(0);
ll ans = data[0].last_y;
while (data[0].pop_last()) {
ans = std::min(ans, data[0].last_y);
}
std::cout << ans << '\n';
return 0;
}
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