Probleme de informatica - enunturi si rezolvari

Harta rutiera a unei tari este formata din n orase si m sosele intre 2 orase, fiecare sosea fiind caractezizata prin distanta in kilometri (numar natural).
Trei prieteni se gasesc in 3 orase distincte x, y, z si isi propun sa se intalneasce intr-un oras t.
Calculati si afisati distanta minima pe care o parcurge fiecare prieten din orasul de pornire pana in orasul t de destinatie.
Afisati cele trei orase de plecare in ordinea in care cei trei prieteni ajung in orasul t.
Exemplu:
date.in
12 19 (n,m)
1 2 20 (oras 1, oras 2, distanta)
1 3 35
1 7 20
2 4 30
3 4 40
3 6 40
3 8 80
4 5 25
5 6 5
6 8 30
6 9 10
7 8 15
7 11 100
8 9 40
8 10 30
8 11 35
9 10 30
10 12 25
11 12 10
1 5 3 12 (x, y, z, t)
date.out
Distantele parcurse de cei trei prieteni:
80 70 105
Orasele de plecare in ordinea in care ajung cei trei prieteni:
5 1 3

#include<fstream>
using namespace std;
const int inf=100000;
int c[200][200],d[200],p[200],n,m,x,y,z,t;
ifstream f("d.in");
ofstream g("d.out");

void citire()
{ int i,j,k,cost;
  f>>n>>m;
  for(i=1;i<=n;i++)
    for(j=1;j<=n;j++)
      if(i==j) c[i][j]=0;
      else c[i][j]=inf;
  for(k=1;k<=m;k++)
    { f>>i>>j>>cost;
      c[i][j]=cost;
      c[j][i]=cost;
    }
  f>>x>>y>>z>>t;
}

void dijkstra(int s)
{ int i,j,k,minn;
  for(i=1;i<=n;i++)
     {
         d[i]=c[s][i];
         p[i]=0;
     }

  p[s]=1;
  for(k=1;k<n;k++)
   { minn=inf;
     for(i=1;i<=n;i++)
       if(!p[i] && d[i]<minn)
                 { minn=d[i]; j=i;
			     }
     for(i=1;i<=n;i++)
       if(!p[i] && d[i]>d[j]+c[j][i])
            d[i]=d[j]+c[j][i];
     p[j]=1;
   }
}

int main()
 { int dx,dy,dz;
   citire();
   dijkstra(x);
   dx=d[t];
   dijkstra(y);
   dy=d[t];
   dijkstra(z);
   dz=d[t];
   g<<"Distantele parcurse de cei trei prieteni:\n";
   g<<dx<<" "<<dy<<" "<<dz<<endl;
   g<<"Ordinea in care ajung cei trei prieteni:\n";
   if(dx<=dy && dy<=dz) g<<x<<" "<<y<<" "<<z;
   else if(dx<=dz && dz<=dy) g<<x<<" "<<z<<" "<<y;
   else if(dy<=dx && dx<=dz) g<<y<<" "<<x<<" "<<z;
   else if(dy<=dz && dz<=dx) g<<y<<" "<<z<<" "<<x;
   else if(dz<=dx && dx<=dy) g<<z<<" "<<x<<" "<<y;
   else if(dz<=dy && dy<=dx) g<<z<<" "<<y<<" "<<x;
   f.close();
   g.close();
   return 0;
}

21 nov 2024

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