POJ1958 Strange Towers of Hanoi [递推]

　　题目传送门

Strange Towers of Hanoi

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 3117		Accepted: 2004

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.

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　　分析：题目大意就是要求你解出n个盘子4座塔的Hanoi问题的最少步数，不需要输入，直接输出n为1-12的所有答案即可。我们知道，一般的三塔Hanoi问题的递推式是d[i]=d[i-1]*2+1，意思就是先将i-1个盘子放在第二个塔上，再把最后一个放在第三个塔上，再将i-1个盘子放在第三个塔上（如果这个不知道就自己去玩一下Hanoi），当然这种方法实质上是将i个盘子的问题先转化为i-1个盘子的问题。那么做这题就可以用类似的思维，先将i个盘子的四塔问题转化为j个盘子的三塔问题（0<=j<=i），令f[i]为i个盘子的四塔问题的答案，则f[i]=min(f[i],f[j]*2+d[i-j])。实际上也就等效于先做j个盘子的四塔问题，再做i-j个盘子的三塔问题，再做一次j个盘子的四塔问题。那么答案就很容易了。

　　Code:

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<iomanip>
#include<algorithm>
#define Fi(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
int d[],f[];
int main()
{
  Fi(i,,)d[i]=d[i-]*+;memset(f,0x3f3f3f3f,sizeof(f));
  f[]=;Fi(i,,)Fi(j,,i)f[i]=min(f[i],*f[j]+d[i-j]);
  Fi(i,,)cout<<f[i]<<endl;return ;
}