Submission #872126

#	Time	Username	Problem	Language	Result	Execution time	Memory
872126		dsyz	Crocodile's Underground City (IOI11_crocodile)	C++17	100 / 100	524 ms	104888 KiB

crocodile

#include <bits/stdc++.h>
#include "crocodile.h"
using namespace std;
using ll = long long;
#define MAXN (100005)
vector<pair<ll,ll> > v[MAXN];
priority_queue<pair<ll,ll>,vector<pair<ll,ll> >,greater<pair<ll,ll> > > pq;
ll secondbest[MAXN]; //second best distance
ll visited[MAXN]; //number of times node i is visited
int travel_plan(int N, int M, int R[][2], int L[], int K, int P[]){
	for(ll i = 0;i < M;i++){
		ll a = R[i][0];
		ll b = R[i][1];
		ll c = L[i];
		v[a].push_back({c,b});
		v[b].push_back({c,a});
	}
	memset(secondbest,-1,sizeof(secondbest));
	for(ll i = 0;i < K;i++){
		pq.push({0,P[i]});
		visited[P[i]] = 2; //we only need to visit exit nodes once anyways so just set it to 2
	}
	while(!pq.empty()){
		ll d = pq.top().first;
		ll x = pq.top().second;
		pq.pop();
		visited[x]++;
		if(visited[x] >= 2){ //if visited at least twice, we can choose the second best distance among all the distances
			if(secondbest[x] != -1 && d >= secondbest[x]) continue; //worse than pre-existing second best distance so far
			if(secondbest[x] == -1){
				secondbest[x] = d;
			}else{
				secondbest[x] = min(secondbest[x],d);
			}
		}else{
			continue;
		}
		for(auto u : v[x]){
			if(secondbest[u.second] == -1 || secondbest[x] + u.first < secondbest[u.second]){
				pq.push({secondbest[x] + u.first,u.second});
			}
		}
	}
	return secondbest[0];
}

• Subtask 1: 46 / 46
• Subtask 2: 43 / 43
• Subtask 3: 11 / 11