UVA - 10795 A Different Task

最新推荐文章于 2021-01-20 21:52:48 发布

acm_1361677193

最新推荐文章于 2021-01-20 21:52:48 发布

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本文链接：https://blog.youkuaiyun.com/acm_1361677193/article/details/40858145

本文介绍了一种变形汉诺塔问题的解决方法，通过动态规划思想实现暴力遍历求解，找出两个任意初始状态间转换所需的最小移动次数。

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A Different Task

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Submit Status

Description

A Different Task

$\epsfbox{p10795a.eps}$

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.

$\epsfbox{p10795b.eps}$

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

Problem setter: Md. Kamruzzaman
Special Thanks: Derek Kisman (Alternate Solution), Shahriar Manzoor (Picture Drawing)

Miguel Revilla 2004-12-10

分析：

变形的汉诺塔问题。感觉就是dp的思想，再简单讲就是暴力了，直接硬遍历。

代码：

#include <iostream>
#include<cstdio>
using namespace std;

long long f(int *p,int i,int final)
{
if(i==0) return 0;
if(p[i]==final) return f(p,i-1,final);
return f(p,i-1,6-p[i]-final)+(1LL<<(i-1));//算术运算符　优先于　位移位运算符优先于关系运算符，所以写成1LL<<(i-1)就错了，因为缺了小括号
//i的变化其实就是在遍历p[i]
}

const int maxn=60+10;
int n,start[maxn],finish[maxn];

int main()
{
int kase=0;
while(scanf("%d",&n)==1&&n)
{
for(int i=1;i<=n;i++)
scanf("%d",&start[i]);
for(int i=1;i<=n;i++)
scanf("%d",&finish[i]);
int k=n;
while(k>=1&&start[k]==finish[k]) k--;
long long ans=0;
if(k>=1)
{
int other=6-start[k]-finish[k];
ans=f(start,k-1,other)+f(finish,k-1,other)+1;
}
printf("Case %d: %lld\n",++kase,ans);
}
return 0;
}