Cod sursa(job #3205109)

Utilizator danutbodbodnariuc danut danutbod Data 18 februarie 2024 19:43:25
Problema Algoritmul lui Dijkstra Scor 100
Compilator cpp-64 Status done
Runda Arhiva educationala Marime 1.97 kb
#include <fstream>
#include <queue>
#include <vector>
#define dim 50002
#define inf 1e9
using namespace std;
ifstream fin("dijkstra.in");
ofstream fout("dijkstra.out");

int n,m;
int d[dim],sel[dim];
vector<pair<int,int>>a[dim];  //a[x] are muchii  (x,y) <y,cost>

struct comp{
    bool operator()(int a,int b){
        return d[a]>d[b];
    }
};

priority_queue<int,vector<int>,comp>Q;   // min heap Q (trebuie 3 parametrila folosirea comp!)>
                                         //un  min heap cu nodurile dupa d[]
void dijkstra(){
    for(int i=2;i<=n;++i) d[i]=inf;
    sel[1]=1;
    Q.push(1);
    while(!Q.empty()){
        int nod=Q.top();
        sel[nod]=0;
        Q.pop();
        for(auto v:a[nod]){
            int vecin=v.first;
            int cost=v.second;
            if(d[vecin]>cost+d[nod]){
               d[vecin]=cost+d[nod];
                if(!sel[vecin]) Q.push(vecin),sel[vecin]=1;
            }
        }
    }
}

int main()
{
    fin>>n>>m;
    for(int i=1;i<=m;i++){
        int x,y,c;
        fin>>x>>y>>c;
        a[x].push_back({y,c});
    }

    dijkstra();

    for(int i=2;i<=n;++i)
        if(d[i]!=inf) fout<<d[i]<<" ";
        else fout<<0<<" ";
    return 0;
}
//afisarea din Dijkstra la fiecare pas!
//       fout<<nod<<": ";for(int i=1;i<=n;i++) if(sel[i])fout<<i<<" ";fout<<" -> ";
//       for(int i=1;i<=n;i++) fout<<d[i]<<" ";fout<<endl;
////in
//5 6
//1 2 1
//1 4 2
//4 3 4
//2 3 2
//4 5 3
//3 5 6
////out
//1: 2 4  -> 0 1 1000000000 2 1000000000
//2: 3 4  -> 0 1 3          2 1000000000
//4: 3 5  -> 0 1 3          2 5
//3: 5    -> 0 1 3          2 5
//5:      -> 0 1 3          2 5
//1 3 2 5


////in
//5 9
//1 2 1
//1 3 9
//4 1 1
//1 5 3
//5 2 4
//2 4 3
//2 3 7
//4 3 2
//5 4 2
////out
//1: 2 3 5  -> 0 1 9 1000000000 3
//2: 3 4 5  -> 0 1 8          4 3
//5: 3 4    -> 0 1 8          4 3
//4: 3      -> 0 1 6          4 3
//3:        -> 0 1 6          4 3
//1 6 4 3