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Curs 8 - Drumuri de cost minim [pdf] - Andrei

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Exemplu (6)<br />

0 3 -1 4 -4<br />

0 1 -3 4 -4<br />

3 0 -4 1 -1<br />

3 0 -4 1 -1<br />

D = 7 4 0 5 3<br />

D = 7 4 0 5 3<br />

2 -1 -5 0<br />

8 5 1 6<br />

-2<br />

0<br />

2<br />

2 -1 -5 0<br />

8 5 1 6<br />

-2<br />

0<br />

p =<br />

nil 1 4 2 1<br />

4 nil 4 2 1<br />

4 3 nil 2 1<br />

4 3 4 nil 1<br />

4 3 4 5 nil<br />

1<br />

-4<br />

3<br />

5<br />

7<br />

8<br />

6<br />

2<br />

1<br />

4<br />

4<br />

Proiectarea Algoritmilor 2010<br />

Închi<strong>de</strong>rea tranzitivă<br />

3<br />

-5<br />

p =<br />

nil 3 4 5 1<br />

4 nil 4 2 1<br />

4 3 nil 2 1<br />

4 3 4 nil 1<br />

4 3 4 5 nil<br />

Fie G = (V,E). Închi<strong>de</strong>rea tranzitivă a lui<br />

EeunG*=(V,E*), E e un G (V,E ), un<strong>de</strong><br />

1, daca i..j<br />

E*(i,j)=<br />

0, daca i..j<br />

Poate fi <strong>de</strong>terminată prin modificarea<br />

algoritmului Floyd-Warshall:<br />

min operatorul boolean sau (˅)<br />

+ operatorul boolean si (˄)<br />

Proiectarea Algoritmilor 2010<br />

5/8/2010<br />

16

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