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POJ1958总结

Strange Towers of Hanoi

题目来源

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!

1958 1

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.

我的代码

这一题没有输入,只有输出,且输出结果是:

1
3
5
9
13
17
25
33
41
49
65
81

其实题设里面已经把解题思路描述出来了:一次四阶的汉诺塔移动是,先将若干个盘子从柱子A用四阶汉诺塔移动的方式移动到中间柱B上,再将A剩下的盘子用三阶汉诺塔的方式移动到目标柱子D上,最后将中间柱B的盘子用四阶汉诺塔的方式移动到目标柱子D上。

要求的是最少需要多少步移动才能完成,那就要遍历出所有可能性,并且选出步数最少的。可以发现第一步和第三步的操作次数都相同。设dp[i]为用四阶汉诺塔方式来移动所需要的最少的操作次数,dp方程是dp[i]=dp[i]<pow[j]+2dp[i-j]?dp[i]:pow[j]+2dp[i-j],i-j就是遍历的所有第一步移动的盘子数量,pow[j]是第二步中将j个盘子用三阶汉诺塔移动所需要的步数,且有公式为(2^i)-1。

#include<stdio.h>

int dp[15];
int pow[15];

int main()
{
	pow[0]=0;
	dp[0]=0;
	for (int i=1;i<13;i++)
	{
		pow[i]=2*pow[i-1]+1;
		dp[i]=pow[i];//dp初始化为标准汉诺塔的移动次数 
	}
	for (int i=1;i<13;i++)
	{
		//j是留在A柱上的 
		for (int j=i-1;j>=1;j--)
		{
			dp[i]=dp[i]<pow[j]+2*dp[i-j]?dp[i]:pow[j]+2*dp[i-j];
		}
		printf("%d\n",dp[i]);
	}
	return 0;
}

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