已知文法G(S):S→aS|bS|c
(1)列出这个文法的所有LR(0)项目
(2)计算项目集规范族并构造识别活前缀的D
（3）构造LR(0)分析表。
(1) LR(0) 项目是指形如 [A → α·β] 的项，其中 A 是文法符号，α 和 β 是文法串。在这个文法 G(S): S → aS | bS | c 中，我们有以下的 LR(0) 项目：
I0:
[S' → ·S]
[S → ·aS]
[S → ·bS]
[S → ·c]
I1:
[S' → S·]
I2:
[S → a·S]
[S → ·aS]
[S → ·bS]
[S → ·c]
I3:
[S → b·S]
[S → ·aS]
[S → ·bS]
[S → ·c]
I4:
[S → c·]
[S → ·aS]
[S → ·bS]
[S → ·c]
(2) 接下来，我们需要构造这个文法的 LR(0) 项集规范族，并且基于这个规范族构造 DFA。构造项集规范族的过程如下：
I0:
[S' → ·S]
[S → ·aS]
[S → ·bS]
[S → ·c]
I1:
[S' → S·]
I2:
[S → a·S]
[S → ·aS]
[S → ·bS]
[S → ·c]
I3:
[S → b·S]
[S → ·aS]
[S → ·bS]
[S → ·c]
I4:
[S → c·]
[S → ·aS]
[S → ·bS]
[S → ·c]
I5:
[S → aS·]
[S → ·aS]
[S → ·bS]
[S → ·c]
I6:
[S → bS·]
[S → ·aS]
[S → ·bS]
[S → ·c]
I7:
[S → aS·]
[S → ·aS]
[S → ·bS]
[S → ·c]
I8:
[S → bS·]
[S → ·aS]
[S → ·bS]
[S → ·c]
I9:
[S → c·]
[S → aS·]
[S → ·aS]
[S → bS·]
[S → ·bS]
[S → ·c]
I10:
[S → aS·]
[S → ·aS]
[S → ·bS]
[S → ·c]
I11:
[S → bS·]
[S → ·aS]
[S → ·bS]
[S → ·c]
I12:
[S → aS·]
[S → ·aS]
[S → ·bS]
[S → ·c]
I13:
[S → bS·]
[S → ·aS]
[S → ·bS]
[S → ·c]
I14:
[S → c·]
[S → aS·]
[S → ·aS]
[S → bS·]
[S → ·bS]
[S → ·c]
在这里，我们可以使用以下的 DFA 来识别输入的活前缀：
+---a--+ +---b--+
| | | |
v v v v
=>(I0)---a-->(I2)---a--->(I5)---ε-->(I10)*
| | |
b b c
| | |
v v v
(I3) (I6) (I4)*
| | |
a a a