Submission #1102645

#	Time	Username	Problem	Language	Result	Execution time	Memory
1102645		asli_bg	Crocodile's Underground City (IOI11_crocodile)	C++11	0 / 100	3 ms	6480 KiB

crocodile

#pragma GCC optimize("O3,unroll-loops")
#include "crocodile.h"
 
#include<bits/stdc++.h>
using namespace std;
 
#define fi first
#define se second
#define all(x) x.begin(),x.end()
#define FOR(i,a) for(int i=0;i<(a);i++)
#define FORE(i,a,b) for(int i=(a);i<(b);i++)
 
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<bool> vb;
typedef vector<pii> vii;
 
#define cont(x) for(auto el:x) {cout<<el<<' ';} cout<<endl;
#define endl '\n'
#define contp(x) for(auto el:x) {cout<<el.fi<<'-'<<el.se<<' ';} cout<<endl;
#define pb push_back
 
#define sp <<' '<<
 
#define DEBUG(x) {cout<<(#x) sp x<<endl;}
 
const int MAXN=1e5+5;
const int INF=1e9;
 
int n,m,k;
vii adj[MAXN];
int mn[MAXN],mn2[MAXN];
bool cikis[MAXN];

void dij(int bas){
    set<pii> s;
    s.insert({0,bas});

    while(!s.empty()){
        auto el=*s.begin();
        s.erase(*s.begin());

        int nd=el.se;
        int dd=el.fi;

        for(auto ell:adj[nd]){
            int kom=ell.fi;
            int cost=ell.se;

            vi temp;
            temp.pb(mn[kom]);
            temp.pb(mn2[kom]);
            temp.pb(dd+cost);
            sort(all(temp));

            mn[kom]=temp[0];

            if(temp[1]!=mn2[kom]){
                s.erase({mn2[kom],kom});
                mn2[kom]=temp[1];
                s.insert({mn2[kom],kom});
            }

        }
    }
}

int travel_plan(int N, int M, int R[][2], int L[], int K, int P[]){
    n=N;
    m=M;
    k=K;
    FOR(i,m){
        int bir=R[i][0];
        int iki=R[i][1];
        bir++;iki++;
        adj[bir].pb({iki,L[i]});
        adj[iki].pb({bir,L[i]});
    }
 
    FOR(i,k){
        P[i]++;
        cikis[P[i]]=true;
    }

    FORE(i,1,n+1){
        if(cikis[i]) mn[i]=mn2[i]=0;
        else mn[i]=mn2[i]=INF;
    }

    FORE(i,1,n+1){
        if(cikis[i]){
            dij(i); //shortest path hesapla
        }
    }

    return mn2[1];
}

• Subtask 1: 0 / 46
• Subtask 2: 0 / 43
• Subtask 3: 0 / 11