| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 996517 | 2024-06-10T18:15:26 Z | 3lektra | 악어의 지하 도시 (IOI11_crocodile) | C++14 | 0 ms | 0 KB |
#include <bits/stdc++.h>
#define vi vector<int>
#define vc vector
#define pi pair<int,int>
using namespace std;
vc<vc<pi>> edg; // Edges, aka the corridors coming out from each room
vc<pi> dis; // The two minimum distances found between 2 nodes. The fastest route will just be used to calculate the second fastest, which is the oke that will be used to obtain the other distances
int fpath = -1e9-7; // Fastest path
int u, c; // Amazing the amount of situations when a node and a cost can be useful
void dijkstra(vi p){ // First Node
priority_queue<pi> tv; // To Visit, not TeleVision or To Vandalize
for(int i : p){
tv.push({0,i});
dis[i].first = 0;}
while(!tv.empty()){
u = tv.top().second; // The node being explored right now
c = tv.top().first; // The cost to get to u
tv.pop();
if(dis[u].second>c){continue;} // If this path is worse then why bother
if(u == 0){
fpath = max(fpath, c);
continue;}
for(pi p : edg[u]){ // p.first = cost, p.second = node the path leads to
if (p.first + c > dis[p.second].second){ // If this is path you've found is fastest than the second fastest one...
if (p.first + c > dis[p.second].first){ // And than the fastest one...
dis[p.second].second = dis[p.second].first; // Then the previously fastest one becomes the second fastest one...
dis[p.second].first = p.first + c;} // This new path becomes the fastest!
else{ // But if it's not fastest than the fastest one...
dis[p.second].second = p.first + c;}
tv.push({dis[p.second].second, p.second});
}
}
}
}
int travel_plan(int n, int m, int (*) 2 r, int l, int k, int p){
edg = vc<vc<pi>>(n);
for(int i = 0; i < m; i++){
edg[r[i][0]].push_back({-l[i],r[i][1]});
edg[r[i][1]].push_back({-l[i],r[i][0]});}
dis = vc<pi>(n,{-1e9-7,-1e9-7});
dijkstra(p);
//for(pi i : dis){cout << i.first << ' ' << i.second << " -- ";}
//cout << -fpath;
return -fpath;}
return 0;}