Transcript Parcurgerea grafurilor orientate

Consideram un graf orientat G=(V,U) cu n varfuri,in care
fiecarui arc, ii asociem un cost (nr. intreg) , semnificatia
acestui cost poate fi de exemplu: costul deplasarii unui
autoturism..;
Matricea costurilor este o matrice A cu n linii si m
coloane care poate avea 2 forme:
A=
c, daca exista arc de cost c>0 de la i la j;
0, daca i=j;
+∞,daca nu exista arc;
A=
c, daca exista arc de cost de la i la j;
0, daca i=j;
-∞, daca nu exista arc de cost de la i la j;
V={(1,2)
u1
u1=7;
u2=3;
(1,4) (2,4) (3,2) (3,4)}
u2 u3 u4 u5
u3=4;
u5=2;
u4=6;
1
A=
0
+∞
+∞
+∞
7
0
6
+∞
+∞
+∞
0
+∞
3
4
2
0
2
3
4
Un drum este minim atunci cand lungimea
lui este cea mai mica si se numeste drum de cost
minim.
Un drum este maxim cand lungimea lui este cea
mai mare si se numeste drum de cost maxim.
Exista 3 posibilitati de a determina drumul
optim:
-sursa sigura, destinatie unica (intre 2 noduri date);
-sursa unica , destinatie multipla (Dijkstra);
-sursa multipla, destinatie multipla (Roy-Floyd);
V={(1,2)(1,4)(2,3)(3,2)(4,3)(5,1)}
u1 u2 u3 u4 u5 u6
u1= 4; u2=6; u3=9;
u4=1 ; u5=7; u6=3;
1
2
5
4
3
0
+∞
A= +∞
+∞
3
A=
0
-∞
-∞
-∞
3
4
0
9
+∞
+∞
4
0
9
-∞
-∞
+∞
1
0
+∞
+∞
-∞
1
0
-∞
-∞
6
+∞
7
0
+∞
+∞
+∞
+∞
+∞
0
6
-∞
7
0
-∞
pentru drumuri
minime
-∞
-∞
-∞
-∞
0
pentru drumuri
maxime
# include<iostream.h>
float a[50][50];
int n;
void drum(int i , int j)
{int k=1, gasit=0;
while ((k<=n) &&(gasit==0))
{if ((i!=k) && (j!=k))
a[i][j]=a[i][k]+a[k][j];
{ drum(i,k);
drum (k,j);
gasit=1;}
k++;{
if(gasit==0)
cout<< j<<“ ”;}
void scrie_drum(int nod_initial, int nod_final)
{ if (a[nod_initial][nod_final]<pinfint)
{cout<<“drumul de la”<<nod_initial<<“la”<<nod_final<<“are
lungimea”<<a[nod_initial][nod_final]<<endl;
cout<<nod_initial<<“ ”;
drum(nod_initial, nod_final);}
else
cout<<“nu exista drum de la”<<nod_initial<<“la”<<nod_final;}
void lungime_drumuri ()
{ int i , j , k;
for ( k=1; k<=n; k++)
for ( i=1; i<=n; i++)
for ( j=1; j<=n; j++)
if ( a[i][j]>a[i][k]+a[k][j] )
a[i][j]=a[i][k]+a[k][j];}
void citire_cost()
{ cout<< “ n= “;
cin >> n ;
for (i=1 ; i<=n; i++ )
for (j=1; j<=n; j++)
cin >>a[i][j]; }
void main ()
{ citire_cost();
lungime_drumuri();
scriu_drum( 5,p);}
# include<iostream.h>
float a[50][50] ,d[50], min ;
int s[50], t[50], n, i, j, r, poz ;
void drum ( int i)
{ if t[i]==1)
drum ( t[i] );
cout<< i<<“ ”; }
void main ()
{ cout<<“dati nr. de noduri ”; cin>> n ;
for(i=1; i<=n; i++)
for (j=1; j<=n; j++)
cin>> a[i] [j] ; }
cout<<“ dati nodul de plecare ”; cin>> r ;
s[ r]=1;
for( i=1; i<=n; i++)
{d [ i]= a[ r][ i] ;
if ( i != r)
if ( d[ i]< pinfinit )
t [ i]= r; }
for (i=1; i<=n-1; i++)
{min= pinfinit ;
for (j=1; j<=n; j++)
if ( s[j ]==0)
if ( d[ j]<min )
{ min = d[ j ] ;
poz=j; }
s [poz]=1;
for (j=1; j<=n; j++)
if (s[ j]==0)
if (d [j]> d [poz]+a[poz][ j])
{ d [ j]=d[poz]+a[poz][ j] ;
t[ j]=poz ; } }
for (i=1; i<=n; i++)
cout<< d[ i]<<“ ”;
cout<<endl;
for (i=1; i<=n; i++)
if (i != r)
if (t[ i]==1)
cout<<“distanta de la”<< r<<“la”<< i<<“este”<< d [ i]
<<endl ;
drum ( i) ;
else
cout<<“nu exista drum de la”<< r<<“ la ”<< i<<endl; }