Uva 10795 A Different Task 解题报告（递归+思维）

本文链接：https://blog.youkuaiyun.com/sf____/article/details/37811061

本文探讨了一种灵活版的汉诺塔问题，不同于传统汉诺塔问题中所有圆盘初始均位于同一柱子上的情形，该问题允许圆盘初始分布在任意柱子上。文章介绍了解决这一问题的方法，并提供了一个示例程序。

A Different Task

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The (Three peg) Tower of Hanoi problem is a popular one in computer science. Briefly the problem is to transfer all the disks from peg-A to peg-C using peg-B as intermediate one in such a way that at no stage a larger disk is above a smaller disk. Normally, we want the minimum number of moves required for this task. The problem is used as an ideal example for learning recursion. It is so well studied that one can find the sequence of moves for smaller number of disks such as 3 or 4. A trivial computer program can find the case of large number of disks also.

Here we have made your task little bit difficult by making the problem more flexible. Here the disks can be in any peg initially.

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If more than one disk is in a certain peg, then they will be in a valid arrangement (larger disk will not be on smaller ones). We will give you two such arrangements of disks. You will have to find out the minimum number of moves, which will transform the first arrangement into the second one. Of course you always have to maintain the constraint that smaller disks must be upon the larger ones.

Input

The input file contains at most 100 test cases. Each test case starts with a positive integer N ( 1 $\le$ N $\le$ 60), which means the number of disks. You will be given the arrangements in next two lines. Each arrangement will be represented by N integers, which are 1, 2 or 3. If the i-th ( 1 $\le$ i $\le$ N) integer is 1, you should consider that i-th disk is on Peg-A. Input is terminated by N= 0. This case should not be processed.

Output

Output of each test case should consist of a line starting with `Case #: ' where # is the test case number. It should be followed by the minimum number of moves as specified in the problem statement.

Sample Input

Sample Output

Case 1: 7
Case 2: 3
Case 3: 0

    解题报告： 《训练指南》例题。很考思维。

    看书上的方法做的。最考思维的地方应该是对于初始态转移到参考态，与参考态转移到目标态。这里直接逆向思维了，由目标态到参考态。

    而中间让时间大大缩短的是，i-1个盘的整体移动，直接使用经典结论了。觉得要好好思考下才能掌握这题的精髓……

    代码如下：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <string>
using namespace std;

#define ff(i, n) for(int i=0;i<(n);i++)
#define fff(i, n, m) for(int i=(n);i<=(m);i++)
#define dff(i, n, m) for(int i=(n);i<=(m);i++)
typedef long long LL;
typedef unsigned long long ULL;
void work();

int main()
{
#ifdef ACM
    freopen("in.txt", "r", stdin);
#endif // ACM

    work();
}

///////////////////////////////////////////
LL ans;
int goal[66];
int now[66];

LL move(int * state, int pan, int pos)
{
    if(pan==0) return 0;
    if(state[pan] == pos)
        return move(state, pan-1, pos);
    return move(state, pan-1, 6-pos-state[pan])+(1LL << (pan-1));
}

void work()
{
    int n;
    int cas = 1;
    while(~scanf("%d", &n) && n)
    {
        ff(i, n) scanf("%d", now+i+1);
        ff(i, n) scanf("%d", goal+i+1);

        while(n>=1 && now[n]==goal[n]) n--;

        LL ans = 0;
        if(n>0)
        {
            int pos;
            if(now[n] == goal[n])
                pos = now[n];
            else
                pos = 6-now[n]-goal[n];
            ans = move(now, n-1, pos)+move(goal, n-1, pos)+1;
        }

        printf("Case %d: %lld\n", cas++, ans);
    }
}