제출 #1021014

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1021014		EkinOnal	악어의 지하 도시 (IOI11_crocodile)	C++17	100 / 100	307 ms	47824 KiB

crocodile

//#pragma GCC optimize("O3,unroll-loops,Ofast")
//#pragma GCC target("avx2,bmi,bmi2,popcnt,lzcnt")
#include <bits/stdc++.h>
//#include <cstdio>
//#include <iostream>
using namespace std;
 
#define MAX 100007
#define pb push_back
//#define mp make_pair 
//#define int long long
#define f first
#define s second
#define vi vector<int>
#define pii pair<int,int>
#define si set<int>
#define vpii vector<pair<int,int>> 
const int mod = 1e9+7;
//const int INF = 1e18;
// myMap.begin()->first :  key
// myMap.begin()->second : value
 
int epow(int a,int b){int ans=1;while(b){if(b&1) ans*=a;a*=a;ans%=mod;a%=mod;b>>=1;}return ans%mod;}
int gcd(int a,int b) {if(a<b)swap(a,b);while(b){int tmp=b;b=a%b;a=tmp;}return a;}
int mul(int a,int b){return ((a%mod)*(b%mod))%mod;}
 
 
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, -1, 0, 1};
 
#define N 50
 
 
 
int solve(int n, int m, int (*edges)[2], int* l, int k, int *entry) 
{	
 
	vector<pii> adj[n+5];
	for(int i=0;i<m;i++)
	{
		adj[edges[i][1]].pb({edges[i][0],l[i]}); adj[edges[i][0]].pb({edges[i][1],l[i]});
	}
 
	vi exits(k);
	for(int i=0;i<k;i++) exits[i]=entry[i];
 
	priority_queue<pii,vector<pii>,greater<pii>> pq;
	vector<priority_queue<int>> dp(n+2);
	vi vis(n+2);
 
	for(int i=0;i<k;i++)
	{
		pq.push({0,exits[i]});
		vis[exits[i]]=1;
		dp[exits[i]].push(0LL); dp[exits[i]].push(0LL);
	}
	
	while(pq.size())
	{
		int node = pq.top().s, cost=pq.top().f;
		pq.pop();
		if(vis[node]>1) continue;
	//	if(vis[node]++) continue;
		vis[node]++;
		if(vis[node]!=2) continue;
		if(node==0LL) return cost;
 
		//if(cost != )
		for(auto u : adj[node])
		{
			if(dp[u.f].size()<2)
			{
				dp[u.f].push(cost+u.s);
				pq.push({cost+u.s,u.f});
			}
			else if(dp[u.f].top() > cost+u.s)
			{
				dp[u.f].pop(); dp[u.f].push(cost+u.s);
				pq.push({cost+u.s,u.f});
			}
		}
	}
	
	cout << "BAD\n";
	
	
    return -1;
}	
 
 
 
int travel_plan(int n, int m, int (*r)[2], int* l, int k, int *p) {
    return solve(n, m, r, l, k, p);
}

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