# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
120934 | Runtime_error_ | Crocodile's Underground City (IOI11_crocodile) | C++14 | 694 ms | 56396 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include "crocodile.h"
#include <bits/stdc++.h>
#include <stdio.h>
#include <stdlib.h>
using namespace std;
const int inf=1e5+9,MX=1e9+9;
vector< pair<int,int> > adj[inf];
bool ex[inf],vis[inf];
//we have to use multi set because we might visit the same node twice with the same distance
//multiset will time out so we use priority queue
priority_queue<pair<int,int> > pq;
int travel_plan(int N, int M, int R[][2], int L[], int K, int P[]){
for(int i=0;i<M;i++)
adj[ R[i][0] ].push_back( make_pair( R[i][1] , L[i] ) ),
adj[ R[i][1] ].push_back(make_pair( R[i][0] , L[i] ));
for(int i=0;i<K;i++)
ex[ P[i] ] = 1,pq.push(make_pair(0, P[i] ));
while(!pq.empty()){
pair<int,int> p = pq.top();
pq.pop();
int u = p.second , dis = -p.first;
if( ex[u] == 0 ){
// we want the second minimum distance so when we reach a node that are not an exit for the first time we abort
//and make that we visited it once
ex[u] = 1;
continue;
}
if(vis[u])
continue;
vis[u] = 1;
if( u == 0)
return dis;
for(auto v:adj[u])
pq.push(make_pair( -(dis+v.second),v.first));
}
}
Compilation message (stderr)
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