【POJ1958】——Strange Towers of Hanoi

 题目翻译起来就四个字——汉诺四塔（原版题面在最后） 

• 首先回顾一下经典汉诺塔，我们设f[i]表示放好i个盘的最少步数。
• 先思考有两个盘子，我们把1盘放到B柱，然后把2盘放到C柱，最后把1盘放到C柱。
• 那么三个盘子就是先执行一遍同样的操作，发现B柱空了。干脆把B柱当作C柱，那么直接把3盘移到B柱上。
• 发现了什么？又变成了两个盘子的情况？！
• 以此类推，i个盘子其实是把执行一次i-1时的步骤，把i盘放到空柱上，最后把剩下i-1个盘子放到i盘所在的柱上。
• 那么递推式显而易见f[i]=f[i-1]+1+f[i-1]=2*f[i-1]+1。

• 思考有4根柱子，我们设ff[i]表示放好i个盘的最少步数。

• 一样思考，我们假设已经放了j（j

• So easy！i-j个盘子和3个柱子。。。？！

• 然后呢？j个盘子和4个柱子

• 那么递推式又双显而易见ff[i]=min{ff[j]+f[i-j]+ff[j]}=min{2*ff[j]+f[i-j]}(1<=j

• 初始化（可以不写）：f[1]=1;ff[1]=1;

结束了，我用的f[0][i]代替f[i]，用f[1][i]代替ff[i]。

#include

using namespace std;

int f[5][25],n=12;

int main()
{
	for(int i=1;i<=n;i++)
	  f[0][i]=0,f[1][i]=2147364847;
	f[0][1]=1;f[1][1]=1;
	for(int i=1;i<=n;i++)
	  f[0][i]=2*f[0][i-1]+1;
	for(int i=2;i<=n;i++)
	  for(int j=1;j

Description

Background 
Charlie Darkbrown sits in another one of those boring Computer Science lessons: At the moment the teacher just explains the standard Tower of Hanoi problem, which bores Charlie to death! 

The teacher points to the blackboard (Fig. 4) and says: "So here is the problem: 

• There are three towers: A, B and C. 
• There are n disks. The number n is constant while working the puzzle. 
• All disks are different in size. 
• The disks are initially stacked on tower A increasing in size from the top to the bottom. 
• The goal of the puzzle is to transfer all of the disks from tower A to tower C. 
• One disk at a time can be moved from the top of a tower either to an empty tower or to a tower with a larger disk on the top.

So your task is to write a program that calculates the smallest number of disk moves necessary to move all the disks from tower A to C." 
Charlie: "This is incredibly boring—everybody knows that this can be solved using a simple recursion.I deny to code something as simple as this!" 
The teacher sighs: "Well, Charlie, let's think about something for you to do: For you there is a fourth tower D. Calculate the smallest number of disk moves to move all the disks from tower A to tower D using all four towers." 
Charlie looks irritated: "Urgh. . . Well, I don't know an optimal algorithm for four towers. . . " 
Problem 
So the real problem is that problem solving does not belong to the things Charlie is good at. Actually, the only thing Charlie is really good at is "sitting next to someone who can do the job". And now guess what — exactly! It is you who is sitting next to Charlie, and he is already glaring at you. 
Luckily, you know that the following algorithm works for n <= 12: At first k >= 1 disks on tower A are fixed and the remaining n-k disks are moved from tower A to tower B using the algorithm for four towers.Then the remaining k disks from tower A are moved to tower D using the algorithm for three towers. At last the n - k disks from tower B are moved to tower D again using the algorithm for four towers (and thereby not moving any of the k disks already on tower D). Do this for all k 2 ∈{1, .... , n} and find the k with the minimal number of moves. 
So for n = 3 and k = 2 you would first move 1 (3-2) disk from tower A to tower B using the algorithm for four towers (one move). Then you would move the remaining two disks from tower A to tower D using the algorithm for three towers (three moves). And the last step would be to move the disk from tower B to tower D using again the algorithm for four towers (another move). Thus the solution for n = 3 and k = 2 is 5 moves. To be sure that this really is the best solution for n = 3 you need to check the other possible values 1 and 3 for k. (But, by the way, 5 is optimal. . . )

Input

There is no input.

Output

For each n (1 <= n <= 12) print a single line containing the minimum number of moves to solve the problem for four towers and n disks.

Sample Input

No input.

Sample Output

REFER TO OUTPUT.

Source

TUD Programming Contest 2002, Darmstadt, Germany