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Submission #998652

# Submission time Handle Problem Language Result Execution time Memory
998652 2024-06-14T12:30:16 Z 3lektra Crocodile's Underground City (IOI11_crocodile) C++14
46 / 100
 2 ms 4696 KB 
#include <bits/stdc++.h>
#include "crocodile.h"
#define vi vector<long long>
#define vc vector
#define pi pair<long long int, long long 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(int k, int p[]){
    priority_queue<pi> tv; // To Visit, not TeleVision or To Vandalize
    for(int i = 0; i < k; i++){
        tv.push({0,p[i]});
        dis[p[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 R[][2], 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(K,P);

return -fpath;}

Subtask #1
46.0 / 46.0

# Verdict Execution time Memory Grader output
1 Correct 1 ms 4444 KB Output is correct
2 Correct 0 ms 4444 KB Output is correct
3 Correct 1 ms 4444 KB Output is correct
4 Correct 1 ms 4444 KB Output is correct
5 Correct 1 ms 4444 KB Output is correct
6 Correct 1 ms 4540 KB Output is correct
7 Correct 1 ms 4440 KB Output is correct
8 Correct 1 ms 4444 KB Output is correct

Subtask #2
0 / 43.0

# Verdict Execution time Memory Grader output
1 Correct 1 ms 4444 KB Output is correct
2 Correct 0 ms 4444 KB Output is correct
3 Correct 1 ms 4444 KB Output is correct
4 Correct 1 ms 4444 KB Output is correct
5 Correct 1 ms 4444 KB Output is correct
6 Correct 1 ms 4540 KB Output is correct
7 Correct 1 ms 4440 KB Output is correct
8 Correct 1 ms 4444 KB Output is correct
9 Incorrect 2 ms 4696 KB Output isn't correct
10 Halted 0 ms 0 KB -

Subtask #3
0 / 11.0

# Verdict Execution time Memory Grader output
1 Correct 1 ms 4444 KB Output is correct
2 Correct 0 ms 4444 KB Output is correct
3 Correct 1 ms 4444 KB Output is correct
4 Correct 1 ms 4444 KB Output is correct
5 Correct 1 ms 4444 KB Output is correct
6 Correct 1 ms 4540 KB Output is correct
7 Correct 1 ms 4440 KB Output is correct
8 Correct 1 ms 4444 KB Output is correct
9 Incorrect 2 ms 4696 KB Output isn't correct
10 Halted 0 ms 0 KB -