图 1 汉诺塔游戏中的柱子和圆盘
图 2 汉诺塔 3 个圆盘的 7 步解决方案
To move n disks from peg 1 to peg 3, using peg 2 as a temporary peg: If n > 0 Then Move n-1 disks from peg 1 to peg 2, using peg 3 as a temporary peg. Move a disk from peg 1 to peg 3. Move n-1 disks from peg 2 to peg 3, using peg 1 as a temporary peg. End If现在来编写一个实现这个解决方案的函数,打印解决游戏的一系列动作。以下代码将使用名称而不是数字来描述柱子。该函数的目标是使用临时柱子(peg3)将一叠圆盘从源柱子(peg1)移动到目标柱子(peg2)。以下是函数的代码:
void moveDisks(int n, string source, string dest, string temp) { if (n > 0) { //将n-1圆盘从源柱子移动到临时柱子 //使用目标柱子作为临时柱子 moveDisks(n - 1, source, temp, dest); //将圆盘从源柱子移动到目标柱子 cout << "Move a disk from " << source << " to " << dest << endl; //将n-1圆盘从临时柱子移动到目标柱子 //使用源柱子作为临时柱子 moveDisks(n - 1, temp, dest, source); } }基本情况在 n = 0 时发生,并且没有要移动的圆盘。在这种情况下,函数调用将不做任何事情返回。下面的程序演示了该函数。
// This program displays a solution to the towers of Hanoi game. #include <iostream> #include <string> using namespace std; // Function prototype void moveDisks(int, string, string, string); int main() { // Play the game with 3 disks moveDisks (3, "peg 1", "peg 3", "peg 2"); cout << "All the disks have been moved!"; return 0; } void moveDisks(int n, string source, string dest, string temp) { if (n > 0) { // Move n-1 disks from source to temp // using dest as the temporary peg moveDisks(n-1, source, temp, dest); // Move a disk from source to dest cout << "Move a disk from " << source << " to " << dest << endl; // Move n-1 disks from temp to dest // using.source as the temporary peg moveDisks(n-1, temp, dest, source); } }程序输出结果:
Move a disk from peg 1 to peg 3
Move a disk from peg 1 to peg 2
Move a disk from peg 3 to peg 2
Move a disk from peg 1 to peg 3
Move a disk from peg 2 to peg 1
Move a disk from peg 2 to peg 3
Move a disk from peg 1 to peg 3
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