Cod sursa(job #2526007)

Utilizator baranceanuBaranceanu Vlad baranceanu Data 18 ianuarie 2020 10:33:28
Problema Arbore partial de cost minim Scor 100
Compilator cpp-64 Status done
Runda Arhiva educationala Marime 1.91 kb
#include <fstream>
#include <queue>

#define NMAX 2000002
using namespace std;
ifstream fin("apm.in");
ofstream fout("apm.out");
int n,m;
int costmin;
int tata[NMAX];
int Find(int x); //returneaza multimea arborelui din care face parte x
int h[NMAX]; ///h[x] inaltimea arborelui cu inaltimea
void citire();
void Kruskal();
void afisare();
void Union(int x,int y);/// reuneste arborele din care face parte x cu arborele din care face y
struct muchie
{
    int x,y,cost;
    friend bool operator >(muchie ,muchie);
};
bool operator >(muchie a,muchie b)
{
    return a.cost>b.cost;
}
muchie sel[NMAX];
priority_queue<muchie,vector<muchie>,greater<muchie> >H;
int main()
{
    citire();
    Kruskal();
    afisare();
    return 0;
}
int Find(int x) //returneaza multimea arborelui din care face parte x
{
    int y;
    int rad=x;
    while (tata[rad])
    {
        rad=tata[rad];
    }
    ///compresez drumul
    while(tata[x])
    {
        y=tata[x];
        tata[x]=rad;
        x=y;
    }
    return rad;
}
void Union(int x,int y)/// reuneste arborele din care face parte x cu arborele din care face y
{
    int rx,ry;
    rx=Find(x);
    ry=Find(y);
    if(h[rx]>h[ry])
    tata[ry]=rx;
    else
        tata[rx]=ry;
    if(h[rx]==h[ry])
        h[ry]++;
}
void citire()
{
    int i;
    muchie a;
    fin>>n>>m;
    for(i=0;i<m;i++)
    {
        fin>>a.x>>a.y>>a.cost;
        H.push(a);
    }
}
void Kruskal()
{
    muchie a;
    int nrsel=0,c1,c2;
    while(nrsel<n-1)
    {
        a=H.top();
        H.pop();
        c1=Find(a.x);
        c2=Find(a.y);
        if(c1!=c2)
        {
            sel[nrsel++]=a;
            costmin+=a.cost;
            Union(c1,c2);
        }
    }
}
void afisare()
{
    int i;
    fout<<costmin<<'\n';
    fout<<n-1<<'\n';
    for(i=0;i<n-1;i++)
    {
        fout<<sel[i].x<<' '<<sel[i].y<<'\n';
    }
}