답안 #998647

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
998647 2024-06-14T12:22:26 Z 3lektra 악어의 지하 도시 (IOI11_crocodile) C++14
0 / 100
1 ms 4444 KB
#include <bits/stdc++.h>
#include "crocodile.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(int p[]){
    priority_queue<pi> tv; // To Visit, not TeleVision or To Vandalize
    for(int i = 0; i < (sizeof(p)/sizeof(int)); 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(P);

return -fpath;}


Compilation message

crocodile.cpp: In function 'void dijkstra(int*)':
crocodile.cpp:15:32: warning: 'sizeof' on array function parameter 'p' will return size of 'int*' [-Wsizeof-array-argument]
   15 |     for(int i = 0; i < (sizeof(p)/sizeof(int)); i++){
      |                               ~^~
crocodile.cpp:13:19: note: declared here
   13 | void dijkstra(int p[]){
      |               ~~~~^~~
crocodile.cpp:15:22: warning: comparison of integer expressions of different signedness: 'int' and 'long unsigned int' [-Wsign-compare]
   15 |     for(int i = 0; i < (sizeof(p)/sizeof(int)); i++){
      |                    ~~^~~~~~~~~~~~~~~~~~~~~~~~~
# 결과 실행 시간 메모리 Grader output
1 Incorrect 1 ms 4444 KB Output isn't correct
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Incorrect 1 ms 4444 KB Output isn't correct
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Incorrect 1 ms 4444 KB Output isn't correct
2 Halted 0 ms 0 KB -