#589
Se dă un graf orientat ponderat cu n noduri și m arce – în care fiecare arc are asociat un cost, număr natural strict pozitiv. Folosind algoritmul Roy-Floyd, construiți matricea drumurilor minime.
| Problema | Roy-Floyd | Operații I/O |
roy-floyd.in/roy-floyd.out
|
|---|---|---|---|
| Limita timp | 0.1 secunde | Limita memorie |
Total: 64 MB
/
Stivă 8 MB
|
| Id soluție | #64644323 | Utilizator | |
| Fișier | roy-floyd.cpp | Dimensiune | 819 B |
| Data încărcării | 20 Mai 2026, 10:21 | Scor/rezultat | Eroare de compilare |
roy-floyd.cpp:3:1: error: ‘ifstream’ does not name a type 3 | ifstream fin("roy-floyd.in"); | ^~~~~~~~ roy-floyd.cpp:4:1: error: ‘ofstream’ does not name a type 4 | ofstream fout("roy-floyd.out"); | ^~~~~~~~ roy-floyd.cpp: In function ‘void citire()’: roy-floyd.cpp:7:5: error: ‘fin’ was not declared in this scope 7 | fin>>n>>m; | ^~~ roy-floyd.cpp: In function ‘void afis()’: roy-floyd.cpp:19:43: error: ‘fout’ was not declared in this scope 19 | for (int j=1;j<=n;j++)if(i==j)fout<<0<<" "; | ^~~~ roy-floyd.cpp:20:39: error: ‘fout’ was not declared in this scope 20 | else if(a[i][j]==maxi)fout<<-1<<" "; | ^~~~ roy-floyd.cpp:21:22: error: ‘fout’ was not declared in this scope 21 | else fout<<a[i][j]<<" "; | ^~~~ roy-floyd.cpp:22:17: error: ‘fout’ was not declared in this scope 22 | fout<<endl;}} | ^~~~ roy-floyd.cpp:22:23: error: ‘endl’ was not declared in this scope 22 | fout<<endl;}} | ^~~~ roy-floyd.cpp:1:1: note: ‘std::endl’ is defined in header ‘<ostream>’; did you forget to ‘#include <ostream>’? +++ |+#include <ostream> 1 | roy-floyd.cpp:17:9: warning: unused variable ‘i’ [-Wunused-variable] 17 | int i, j; | ^ roy-floyd.cpp:17:12: warning: unused variable ‘j’ [-Wunused-variable] 17 | int i, j; | ^
www.pbinfo.ro permite evaluarea a două tipuri de probleme:
Problema Roy-Floyd face parte din prima categorie. Soluția propusă de tine va fi evaluată astfel:
Suma punctajelor acordate pe testele utilizate pentru verificare este 100. Astfel, soluția ta poate obține cel mult 100 de puncte, caz în care se poate considera corectă.