共有n個結點,結點編號1~n,設C是G的成本鄰接矩陣,用Dk(i,j)表示從i到j并且不經(jīng)過編號比k還大的結點的最短路徑的長度(Dn(i,j)即為圖G中結點i到j的最短路徑長度),則求解該問題的遞推關系式為()。A、Dk(i,j)=Dk-1(i,j)+C(i,j)B、Dk(i,j)=min{Dk-1(i,j),Dk-1(i,j)+C(i,j)}C、..." />
利用動態(tài)規(guī)劃方法求解每對結點之間的最短路徑問題(all pairs shortest path problem)時,設有向圖G=<V,E>
A、Dk(i,j)=Dk-1(i,j)+C(i,j)
B、Dk(i,j)=min{Dk-1(i,j),Dk-1(i,j)+C(i,j)}
C、Dk (i,j)=Dk-1(i,k)+Dk-1(k,j)
D、Dk(i,j)=min{Dk-1(i,j),Dk-1(i,k)+Dk-1(k,j)}