poj 1958 Strange Towers of Hanoi(递推)

Strange Towers of Hanoi

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 2305		Accepted: 1524

DescriptionBackgroundCharlie 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. . . "ProblemSo 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 toCharlie, 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 movedto 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 findthe 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 thelast 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 possiblevalues 1 and 3 for k. (But, by the way, 5 is optimal. . . )InputThere is no input.OutputFor 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.

题意：将汉诺塔的三根柱子改为四根。问将n个盘子从A柱移动到D柱所需要的最少移动次数。

思路：递推。设f表示将n个盘子按照四根汉诺塔的算法从A柱移动到D柱所需要的最少移动次数。容易算得f[1]=1,f[2]=3,f[3]=5。我们知道按照三根汉诺塔的算法最少移动次数为2^n-1。

步骤分三步：

（1）将k（k<n）个盘子按照四根汉诺塔的算法从A柱移动到B柱。因为k<n,所以移动次数在前面已算出，为f[k]。

（2）将n-k个盘子按照三根汉诺塔的算法从A柱移动到D柱。这时移动次数为2^(n-k)-1。

（3）将k个盘子按照四根汉诺塔的算法从B柱移动到D柱。这时移动次数和第（1）步相同。

最终移动次数为三个步骤的移动次数总和。枚举k的值，执行上述步骤。找最小的移动次数总和。

AC代码：

#include <iostream>#include <cstring>#include <string>#include <cstdio>#include <algorithm>#include <queue>#include <cmath>#include <vector>#include <cstdlib>using namespace std;int f[15]={0,1,3,5};int hanio(int k,int m){int sum=f[k]*2;sum+=pow((double)2,(double)m)-1;return sum;}int main(){for(int i=4;i<=12;i++){f[i]=1000000000;for(int j=1;j<i;j++){int t=hanio(j,i-j);f[i]=min(f[i],t);}}for(int i=1;i<=12;i++)cout<<f[i]<<endl;return 0;}

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