# | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
---|---|---|---|---|---|---|---|
906318 | vjudge1 | Job Scheduling (IOI19_job) | C++17 | 64 ms | 15956 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "job.h"
#include <vector>
#pragma GCC target ("avx2")
#pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")
#include<bits/stdc++.h>
#include<math.h>
using namespace std;
typedef long long int ll;
typedef long double ld;
typedef pair<ll, ll> pl;
typedef vector<ll> vl;
#define FD(i, r, l) for(ll i = r; i > (l); --i)
#define K first
#define V second
#define G(x) ll x; cin >> x;
#define GD(x) ld x; cin >> x;
#define GS(s) string s; cin >> s;
#define EX(x) { cout << x << '\n'; exit(0); }
#define A(a) (a).begin(), (a).end()
#define F(i, l, r) for (ll i = l; i < (r); ++i)
#define NN
#define M 1000000007 // 998244353
vl adj[200010];
long long scheduling_cost(std::vector<int> p, std::vector<int> u, std::vector<int> d) {
ll ans = 0;
int n = (int)p.size();
bool case1 = true,case2 = true;
for(int i=1;i<n;i++){
case1 &= p[i]==i-1;
case2 &= p[i]==0;
case2 &= d[i]==1;
adj[p[i]].push_back(i);
}
if(case1){
ll time = 0;
for(int i=0;i<n;i++){
time += d[i];
ans += time*u[i];
}
return ans;
}
if (n <= 200) {
ll ans = 0;
auto cmp = [&](int i,int j){
return -u[i]*d[j] < -u[j]*d[i];
};
auto dfs = [&](auto &&self, ll i) -> vl {
vector<vl> ord;
for (auto x: adj[i]) ord.push_back(self(self, x));
vl ans;
while (ord.size()) {
ll mnidx = 0;
F(i, 1, ord.size()) if (cmp(ord[i][0], ord[mnidx][0])) mnidx = i;
ans.push_back(ord[mnidx][0]);
ord[mnidx].erase(ord[mnidx].begin());
if (!ord[mnidx].size()) ord.erase(ord.begin() + mnidx);
}
ans.insert(ans.begin(), i);
};
ll time = 0;
for (auto i: dfs(dfs, 0)) {
time += d[i]; ans += time * u[i];
}
return ans;
}
auto cmp = [&](int i,int j){
return -u[i]*d[j] > -u[j]*d[i];
};
priority_queue<ll, vector<ll>, decltype(cmp)> pq(cmp);
ans += d[0] * u[0];
ll time = d[0];
for (auto x: adj[0]) pq.push(x);
while (pq.size()) {
auto i = pq.top(); pq.pop();
time += d[i];
ans += time * u[i];
for (auto y: adj[i]) pq.push(y);
}
return ans;
}
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