poj 1958 Strange Towers of Hanoi

原创
2014/11/29 11:42
阅读数 133
Strange Towers of Hanoi
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 2513 Accepted: 1667

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


汉诺塔的解法是一个递归,假设有ABC三个柱子,A柱上有n个盘子。将A上盘子移到C,相当于先把A上n-1个挪到B,再把A上剩余的一个挪到C,现在问题变为了将B上n-1个盘子挪到C。所以有递推公式f(n)=2*f(n-1)+1。
题目中的是加了一根柱子,根据题目的提示,有递推公式g(n)=min{2*g(n-k)+f(k)},即g(n)=min{2*g(n-k)+2^k-1}。可以用dp来做。

#include <cstdio>
#include <cstdlib>
#define MAXN 15
#define INF 100000000

int main()
{
	int dp[MAXN] = {0};
	dp[1] = 1;
	dp[0] = 0;

	for (int i = 2; i <= 12; ++i)
	{
		int my_min = INF;
		for (int j = 1; j <= i; ++j)
		{
			int tmp = (dp[i - j] << 1) + (1 << j) - 1;
			if (my_min > tmp)
			{
				my_min = tmp;
			}
		}
		dp[i] = my_min;
	}

	for (int i = 1; i <= 12; ++i)
	{
		printf("%d\n", dp[i]);
	}
	return 0;
}



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