Cod sursa(job #3342929)

Utilizator alexandraaa_bogdanAlexandra Bogdan alexandraaa_bogdan Data 26 februarie 2026 10:32:49
Problema Arbore partial de cost minim Scor 100
Compilator cpp-64 Status done
Runda Arhiva educationala Marime 1.96 kb
#include <fstream>
#include <queue>
#include <algorithm>
#define NMAX 200002
using namespace std;
ifstream fin("apm.in");
ofstream fout("apm.out");
int n,m;
int tata[NMAX];
int h[NMAX]; //inalt arborelui cu rad x
int costmin;
struct muchie
{
    int x,y,c;
    friend bool operator < (muchie m1,muchie m2);
};
bool operator < (muchie m1,muchie m2)
{
    return m1.c > m2.c;
}
muchie apm[NMAX];
void citire();
void kruskal();
void afisare();
int Find(int x);
void Union(int x, int y);
priority_queue<muchie> H;
int main()
{
    citire();
    kruskal();
    afisare();
    return 0;
}
void Union(int x, int y)
{
    int rx,ry;
    //reuneste arborele in care se afla x cu arborele in care se afla y
    rx=Find(x); ry=Find(y);
    if (h[rx]<h[ry])  tata[rx]=ry; //rx devine fiul lui ry
        else if (h[ry]<h[rx]) tata[ry]=rx; //ry devine fiul lui rx
                 else
                 {
                     tata[ry]=rx;
                     h[rx]++;
                 }
}
int Find(int x)
{
    //determin rad arborelui
    int rx=x,y;
    while (tata[rx]!=0) rx=tata[rx];
    //compresez drumul de la x la rx; fiecare nod de pe acest drum devine fiu al lui rx
    while(tata[x] && tata[x]!=rx)
    {
        y=tata[x];
        tata[x]=rx;
        x=y;
    }
    return rx;
}
void citire()
{
    int i,x,y,c;
    muchie aux;
    fin>>n>>m;
    for (i=1; i<=m; i++)
    {
        fin>>aux.x>>aux.y>>aux.c;
        H.push(aux);
    }
}
void kruskal()
{
    int nrsel,i,j,mic,mare;
    int rx,ry;
    muchie aux;
    nrsel=0; i=1;
    while (nrsel<n-1)
    {
        aux=H.top(); H.pop();
        rx=Find(aux.x); ry=Find(aux.y);
        if (rx!=ry)
        {
            costmin+=aux.c;
            apm[++nrsel]=aux;
            Union(rx,ry);
        }
    }
}
void afisare()
{
    int i;
    fout<<costmin<<'\n';
    fout<<n-1<<'\n';
    for (i=1; i<n; i++)
         fout<<apm[i].x<<' '<<apm[i].y<<'\n';
}