Acdream 1219 The Towers of Hanoi Revisited（递归汉诺塔问题）

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The Towers of Hanoi Revisited

Special Judge Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)

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Problem Description

  You all must know the puzzle named "The Towers of Hanoi". The puzzle has three pegs and N discs of different radii, initially all disks are located on the first peg, ordered by their radii - the largest at the bottom, the smallest at the top. In a turn you may take the topmost disc from any peg and move it to another peg, the only rule says that you may not place the disc atop any smaller disk. The problem is to move all disks to the last peg making the smallest possible number of moves.
  There is the legend that somewhere in Tibet there is a monastery where monks tirelessly move disks from peg to peg solving the puzzle for 64 discs. The legend says that when they finish, the end of the world would come. Since it is well known that to solve the puzzle you need to make 2N - 1 moves, a small calculation shows that the world seems to be a quite safe place for a while.
  However, recent archeologists discoveries have shown that the things can be a bit worse. The manuscript found in Tibet mountains says that the puzzle the monks are solving has not 3 but M pegs. This is the problem, because when increasing the number of pegs, the number of moves needed to move all discs from the first peg to the last one following the rules described, decreases dramatically. Calculate how many moves one needs to move N discs from the first peg to the last one when the puzzle has M pegs and provide the scenario for moving the discs.

Input

  Input file contains N and M (1 ≤ N ≤ 64, 4 ≤ M ≤ 65).

Output

  On the first line output L - the number of moves needed to solve the puzzle. Next L lines must contain the moves themselves. For each move print the line of the form

move from to

if the disc is moved to the empty peg or

move from to atop

if the disc is moved atop some other disc.

Disc radii are integer numbers from 1 to N, pegs are numbered from 1 to M.

Sample Input

5 4

Sample Output

13

move 1 from 1 to 3

move 2 from 1 to 2

move 1 from 3 to 2 atop 2

move 3 from 1 to 4

move 4 from 1 to 3

move 3 from 4 to 3 atop 4

move 5 from 1 to 4

move 3 from 3 to 1

move 4 from 3 to 4 atop 5

move 3 from 1 to 4 atop 4

move 1 from 2 to 1

move 2 from 2 to 4 atop 3

move 1 from 1 to 4 atop 2

题目大意：

就是给你N个盘子，M个柱子，让你求的是怎么样操作使第一个柱子上的全部的盘子移动到最后一个盘子，操作数目最少并将移动的方法写出来。

解题思路：

在这路首先介绍一个算法—Frame算法(这是四柱汉诺塔问题)：

（1）用4柱汉诺塔算法把A柱上部分的n- r个碟子通过C柱和D柱移到B柱上【F( n- r )步】。

（2）用3柱汉诺塔经典算法把A柱上剩余的r个碟子通过C柱移到D柱上【2^r-1步】。

（3）用4柱汉诺塔算法把B柱上的n-r个碟子通过A柱和C柱移到D柱上【F(n-r)步】。

（4）依据上边规则求出所有r（1≤r≤n）情况下步数f(n)，取最小值得最终解。

因此Frame算法的递归方程如下：

F(n)=min(2*F(n-r)+2^r-1)，（1≤r≤n）。

通过这个方程我们能得到所有4柱汉诺塔的步骤个数，同时也有人证明[1]了，对于四柱汉诺塔，当

r=((8?n+1)????????√?1)/2

时，能保证f(n)取得最小值

F(n)=(n?(r2?r+2)/2)?2r+1

。所以算法的复杂度是

F(n)=O((2?n)?????√?2(2?n)√)

。从这这个方程中也可以看出，在n<6的时候，我们可以验证是和我们起初的构想的结构是相同的，但是当n再增多时就不是当初想的那样了。

接下来就是根据四柱的Frame算法来求解M柱的问题

（1）用M柱汉诺塔算法把1柱上部分的n-r个碟子通过3…M柱移到2柱上【M( n- r )步】。

（2）用M-1柱汉诺塔算法把1柱上剩余的r个碟子通过3…M-1柱移到M柱上【(r)步】。

（3）用M柱汉诺塔算法把2柱上的n-r个碟子通过1柱和3…M柱移到M柱上【M( n- r )步】。

（4）依据上边规则求出所有r（1≤r≤n）情况下步数m(n)，取最小值得最终解M(n)。

其实我们的这个问题就是基于这个算法求解的就是一个递归的问题：

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
using namespace std;
const int MAXN = 1e2+5;
const double eps = 1e-9;
const int INF = 1e8+5;
int f[MAXN][MAXN], p[MAXN][MAXN];///f:步数 p:节点
void get(int n, int k)
{
    if(f[n][k] != -1)
        return;
    f[n][k] = INF;
    if(k < 3)
        return;
    for(int m=1; m<n; m++)
    {
        get(m, k);
        get(n-m, k-1);
        int tp = 2*f[m][k]+f[n-m][k-1];
        if(f[n][k] > tp)
        {
            f[n][k] = tp;
            p[n][k] = m;
        }
    }
}
int n, m;
int hanoi[MAXN][MAXN], num[MAXN];
void print(int s, int t, int a, int b)
{
    if(a == 1)
    {
        printf("move %d from %d to %d ",hanoi[s][num[s]]+1,s,t);
        if(num[t])
            printf("atop %d",hanoi[t][num[t]]+1);
        puts("");
        num[t]++;
        hanoi[t][num[t]]=hanoi[s][num[s]--];
        return;
    }
    for(int i=1; i<=m; i++)
    {
        if(i!=s && i!=t)
        {
            if(hanoi[i][num[i]] > hanoi[s][num[s]-p[a][b]+1])
            {
                print(s, i, p[a][b], b);
                print(s, t, a-p[a][b], b-1);
                print(i, t, p[a][b], b);
                return;
            }
        }
    }
    return ;
}
int main()
{
    while(cin>>n>>m)
    {
        memset(f, -1, sizeof(f));
        for(int i=1; i<=m; i++)
            f[1][i] = 1;
        get(n, m);
        cout<<f[n][m]<<endl;
        memset(hanoi, 0, sizeof(hanoi));
        memset(num, 0, sizeof(num));
        for(int i=n; i>=1; i--)
        {
            hanoi[1][num[1]] = i;
            num[1]++;
        }
        for(int i=1; i<=m; i++)
            hanoi[i][0] = INF;
        print(1, m, n, m);
    }
    return 0;
}