| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 882452 | vjudge1 | Jakarta Skyscrapers (APIO15_skyscraper) | C++17 | 0 ms | 0 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,sse4.2")
#include <bits/stdc++.h>
using namespace std;
#define int int64_t
const int MAXM = 3e5+5;
int n, p[MAXM], B[MAXM], m;
struct quad {
int first, second, third, fourth;
quad() : first(0), second(0), third(0), fourth(0) {}
quad(int a, int b, int c, int d) :first(a), second(b), third(c), fourth(d) {}
inline bool operator< (quad other) const {
if(other.first != this->first) return this->first < other.first;
if(other.second != this->second) return this->second < other.second;
if(other.third != this->third) return this->third < other.third;
return this->fourth < other.fourth;
}
inline bool operator> (quad other) const {
if(other.first != this->first) return this->first > other.first;
if(other.second != this->second) return this->second > other.second;
if(other.third != this->third) return this->third > other.third;
return this->fourth > other.fourth;
}
};
signed main() {
cin.tie(0)->sync_with_stdio(0);
cin >> n >> m;
for(int i = 0; i < m; i++) cin >> B[i] >> p[i];
vector<int> a(n, INT_MAX);
vector<vector<int>> b(n);
priority_queue<quad, vector<quad>, greater<quad>> pq;
pq.push({0, B[0], p[0], 3});
for(auto x:b[B[0]]) if(x!=0) pq.push({0, B[0], p[x], 3});
a[B[0]] = min(a[B[0]], 0ll);
for(int i = 0; i < m; i++) b[B[i]].push_back(i);
while(!pq.empty()) {
auto cur = pq.top(); pq.pop();
if(a[cur.second] == INT_MAX) {
for(auto x:b[cur.second]) pq.push({cur.first, cur.second, p[x], 3});
}
a[cur.second] = min(a[cur.second], cur.first);
if(cur.second >= cur.third && cur.fourth != 2)
pq.push({cur.first + 1, cur.second - cur.third, cur.third, 1});
if(cur.second < n - cur.third && cur.fourth != 1)
pq.push({cur.first + 1, cur.second + cur.third, cur.third, 2});
}
cout << (a[B[1]] == INT_MAX ? -1 : a[B[1]]) << '\n';
}