這是資料結構第三章 odp 檔，stack and queue。

1. Chapter 3 Stacks and Queues 3.1 Templates in C++ 3.2 The Stack Abstract Data Type 3.3 The Queue Abstract Data Type 3.4 Subtyping and Inheritance in C++ 3.5 A Mazing Problem 3.6 Evaluation of Expressions 3.7 Multiple Stacks and Queues

3. A template may be viewed as a variable that can be instantiated to any data type, irrespective of whether this data type is a fundamental C++ type or a user-defined type.

4. template 又稱為 parameterized type.

5. 3.1 Templates in C++ template < class KeyType> void sort(KeyType *a, int n) { // 作非遞減排序 for ( int i = 0; i < n; i++) { int j = i; for ( int k = i + 1; k < n; k++) // 找出最小元素位置 if (a[k] < a[ j]) j = k; KeyType temp = a[i]; a[i] = a[j]; a[j] = temp; } } // Program 3.1 Selection sort using templates

7. 若 KeyType 是基本資料結構，則如此寫已足夠；若是 KeyType 是程式設計者自訂的資料結構，就需要針對 < 作運算子重載，也需要對 = 作運算子重載或給 copy constructor 。

9. 3.1 Templates in C++ template < class Type> Bag<Type>::Bag( int MaxBagSize): MaxSize(MaxBagSize) { array = new Type[MaxSize]; top = -1; } template < class Type> Bag<Type>::~Bag() { delete [] array; } template < class Type> void Bag<Type>::Add( const Type& x) { if (IsFull()) Full(); else array[++top] = x; } // 定義在類別外面的成員函式，每一個函式都加上 template <class Type> 宣告方法 Bag < int > a(3); Bag <Rectangle> r(10);

11. Stack : Last-In-First-Out(LIFO) list

12. A stack is an ordered list in which insertions and deletions are made at one end called the top . 例如：插入 A,B,C,D ，刪除一個元素。

13. 令 S = (a 0 , a 1 , ..., a n-1 ) ， a 0 is bottom, a n-1 is top, a i is on top of a i-1 , 0 < i < n.

15. 3.2 The Stack Abstract Data Type template < class KeyType> class Stack { // objects: A finite ordered list of zero or more elements public : Stack ( int MaxStackSize = DefaultSize); Boolean IsFull(); // return TRUE if Stack is full; FALSE otherwise Boolean IsEmpty(); // return TRUE if Stack is empty; FALSE otherwise void Add(const KeyType &); // if IsFull(), return 0; // else insert an element to the top of the Stack KeyType *Delete( KeyType &); // if IsEmpty(), then return 0; // else remove and return a pointer to the top element }; // Abstract data type Stack ， member function 的部份

16. 3.2 The Stack Abstract Data Type template < class KeyType> class Stack{ // data member 的部份 ... private : int top; KeyType *stack; int MaxSize; }; template < class KeyType> // 建構子 Stack <KeyType> :: Stack ( int MaxStackSize):MaxSize(MaxStackSize) { stack = new KeyType [MaxSize]; top = -1; }

17. 3.2 The Stack Abstract Data Type template < class KeyType > inline Boolean Stack< KeyType >:: IsFull (){ if (top == MaxSize – 1) return TRUE; else return FALSE; } template < class KeyType > inline Boolean Stack< KeyType >:: IsEmpty (){ if (top == -1) return TRUE; else return FALSE; }

18. 3.2 The Stack Abstract Data Type template < class KeyType> void Stack< KeyType >:: Add ( const KeyType& x ){ // add x to stack if ( IsFull() ) StackFull(); else stack[++top] = x ; } template < class KeyType> KeyType* Stack<Keytype>:: Delete (KeyType& x) { // Remove top element from stack if (IsEmpty() ) { StackEmpty(); return 0;} x = stack[top--]; return ( & x ); }

19. Exercises 3.2 2. Consider the railroad switching network given in Figure 3.3. Raidroad cars numbered 1, 2, 3, ..., n are initially in the top right track segment (in this order, left to right). Railroad cars can be moved into the vertical track segment one at a time from either of the horizontal segments and then moved from the vertical segment to any one of the horizontal segments. The vertical segment operates as a stack as new cars enter at the top and cars depart the vertical segment from the top. For instance, if n = 3, we could move car 1 into the vertical segment, move 2 in, move 3 in, and then take the cars out producing the new order 3, 2, 1. For n = 3 and 4 what are the possible permutations of the cars that can be obtained? Are any permutations not possible ?

21. First-In-First-Out (FIFO)

23. a 0 is the front element, a n-1 is the rear element,

24. a i is behind a i-1 , 1 ≤ i ≤ n. B C D E A f r r r r r f

29. 3.3 The Queue Abstract Data Type Figure 3.5 Insertion and deletion from a queue front rear Q[0] [1] [2] [3] [4] [5] [6]... Comments -1 -1 queue empty initial -1 0 J1 J1 joins Q -1 1 J1 J2 J2 joins Q -1 2 J1 J2 J3 J3 joins Q 0 2 J2 J3 J1 leaves Q 0 3 J2 J3 J4 J4 joins Q 1 3 J3 J4 J2 leaves Q front rear Q[0] [1] [2] ...... [n-1] Next Operation -1 n-1 J1 J2 J3 ...... Jn initial state 0 n-1 J2 J3 ...... Jn delete J1 -1 n-1 J2 J3 J4 ...... Jn+1 add Jn+1 (J2 through Jn are moved) 0 n-1 J3 J4 ...... Jn+1 delete J2 -1 n-1 J3 J4 J5 ...... Jn+2 add Jn+2 Figure 3.6 Queue example

34. A data object of Type B IS- A data object of Type A if B is more specialized than A or A is more general than B. The set of all objects of Type B is a subset of the set of all objects of Type A.

35. 例如：椅子是家具、獅子是哺乳動物、矩形是多邊形

36. Stack IS-A Bag or Stack is a subtype of Bag.

37. C++ provides a mechanism to express the IS-A relationship called public inheritance .

38. class Stack: public Bag { // Bag 是 base class, Stack 是 derived class Stack( int MaxSize = DefaultSize); ~Stack(); int * Delete( int &); }

39. 3.4 Subtyping and Inheritance in C++ class Bag { public : Bag( int MaxSize = DefaultSize); virtual ~Bag(); virtual void Add( int ); virtual int * Delete( int &); virtual Boolean IsFull(); virtual Boolean IsEmpty(); protected : virtual void Full(); virtual void Empty(); int *array, MaxSize, top; }; class Stack: public Bag { Stack( int MaxSize = DefaultSize); ~Stack(); int * Delete( int &); }

41. Another important consequence of public inheritance is that inherited members have the same level of access in the derived class as they did in the base class. Stack::Stack ( int MaxStackSize) :Bag(MaxStackSize) { } Stack::~Stack() { } // 直接呼叫 Bag 的解構子 int * Stack::Delete( int & x) { if (IsEmpty()) {Empty(); return 0;} x = array[top--]; return &x; }

44. 多訓練幾次，可算出老鼠的學習曲線。

45. 可用程式來模擬，只是理論上，程式應該第二次就可以將完全正確的路徑記起來。 entrance exit

47. maze[i][j] = 0 代表可走的路徑

48. maze[i][j] = 1 代表不可走的路徑 , blocked path

50. 3.5 Amazing Problem X 西北 NW 北 N 東北 NE 東 E 東南 S E 南 S 西南 SW 西 W [ i – 1 ][ j – 1 ] [ i - 1 ][ j ] [ i - 1 ][ j + 1 ] [ i ][ j + 1 ] [ i + 1 ][ j + 1 ] [ i + 1 ][ j ] [ i + 1 ][ j – 1 ] [ i ][ j – 1 ] [ i ][ j ]

51. 3.5 Amazing Problem q move[q].a move[q].b N -1 0 NE -1 1 E 0 1 SE 1 1 S 1 0 SW 1 -1 W 0 -1 NW -1 -1

53. 可以用 maze[m+2][p+2] 的陣列來模擬迷宮，最外圈全部都設為 1 ，這樣就不用考慮邊緣的情形。

55. 3.5 Amazing Problem initialize stack to the maze entrance coordinates and direction east; while ( stack is not empty) { ( i , j , dir ) = coordinates and direction deleted from top of stack ; while (there are more moves) { ( g , h ) = coordinates of next move; if (( g == m ) && ( h == p )) success; if (! maze [ g ][ h ]) && (! mark [ g ][ h ]) { // 有路走且尚未走過 mark [ g ][ h ] = 1; dir = next direction to try; add( i , j , dir ) to top of stack ; i = g; j = h; dir = north ; } } } cout << &quot;no path found&quot; << endl;

57. 用 mark[m+2][p+2] 來記錄是否走過， initialize 為 0 ，走過就設為 1

58. maze[1][1] 與 maze[m][p] 都為 0 。

59. 要將前面的演算法實作出來，還需要一個 stack ，這個 stack 最多有 mp 個元素 ( 最長路徑的長度，每個位置都放到 stack 中 ) ，實際上可能沒那麼多，但可能到 m/2 (p – 2) + 2

60. Stack 的每一項可定義如下： struct items { int x, y, dir; };

61. void path( int m, int p) { mark[1][1] = 1; Stack<items> stack(m*p); items temp; temp.x = 1; temp.y = 1; temp.dir = E; stack.Add(temp); while (!stack.IsEmpty()) { temp = *stack.Delete(temp); int i = temp.x; int j = temp.y; int d = temp.dir; while (d < 8) { int g = i + move[d].a; int h = j + move[d].b; if ((g == m) && (h == p)) { cout << stack; cout << i << &quot; &quot; << j << endl; cout << m << &quot; &quot; << p << endl; return ; } if ((!maze[g][h]) && (!mark[g][h])) { mark[g][h] = 1; temp.x = i; temp.y = j; temp.dir = d + 1; stack.Add(temp); i = g; j = h; d = N; } else d++; } } cout << &quot;no path in maze&quot; << endl; }

63. 裡面的 while 迴圈最多跑八次，每次都是 constant time ，因此整個 while 迴圈是 constant time 。

64. 外面的 while 迴圈最多就是每個點都放到 stack 一次，因此 time complexity 就是 O(mp) 。

65. 真正作老鼠走迷宮的實驗時，可以在出口的地方放 cheese ，用以吸引老鼠跑快一點，用電腦模擬如何將這種變因考慮在內呢？有辦法找到最短路徑嗎？

66. 參考習題： 2. What is the maximum path length from start to finish for any maze of dimensions m * p?

68. Operators: +, –, *, /, %, >, < , ==, >=, <=, !=, &&, ||, !, ...

69. Delimiter: 如 (, )

74. Postfix: A B / C – D E * + A C * –

78. 2.Move all operators so that they replace their corresponding right parentheses.

80. 將每個括號裡的部份都改成 postfix ，即為 AB/ C– DE* + AC* –

84. Behaves as an operator with low priority when it is in stack, in-stack priority ( isp ) = 8

86. isp('#') = icp('#') = 8

87. 將 infix 轉成 postfix 的時間複雜度為 O(n) ， n 為 token 數。

88. void postfix(expression e) { Stack<token> stack; token y; stack.Add('#'); for (token x = NextToken(e); x != '#'; x = NextToken(e)) { if (x is an operand) cout << x; else if (x == ')' ) for (y = *stack.Delete(y); y != '('; y = *stack.Delete(y)) cout << y; else { for (y = *stack.Delete(y); isp(y) <= icp(x); y = *stack.Delete(y)) cout << y; stack.Add(y); stack.Add(x); } } while (!stack.IsEmpty()) cout << *stack.Delete(y); }

90. – A + B – C + D

91. A * –B + C

92. (A + B) * D + E / (F + A * D) + C

93. A && B || C || ! (E > F) // 用 C++ 的 precedence

94. !(A && ! ((B < C) || (C > D))) || (C < E)

96. Write an algorithm to evaluate a prefix expression e. (hint: Scan e right to left and assume that the leftmost token of e is '# ')

97. Write an algorithm to transform an infix expression e into its prefix equivalent. Assume that the input expression e begins with a '# ' and that the prefix expression should begin with a '# '.

99. 若有 n 個 stacks 放在同一個陣列，就只好將 M[m] 切成 n 等份，如下圖。 M 0  m/n – 1   2m/n – 1  m – 1 b[0] b[1] b[2] b[n] t[0] t[1] t[2]

101. 3.7 Multiple Stacks and Queues b[0] t[0] b[1] t[1] b[i] t[i] b[i+1] t[i+1] b[i+2] t[j] b[j+1] b[n]