codeforces 392 B. Tower of Hanoi（汉诺塔 DP）

题目链接：http://codeforces.com/contest/392/problem/B

B. Tower of Hanoi

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The Tower of Hanoi is a well-known mathematical puzzle. It consists of three rods, and a number of disks of different sizes which can slide onto any rod. The puzzle starts with the disks in a neat stack in ascending order of size on one rod, the smallest at the top, thus making a conical shape.

The objective of the puzzle is to move the entire stack to another rod, obeying the following simple rules:

1. Only one disk can be moved at a time.
2. Each move consists of taking the upper disk from one of the stacks and placing it on top of another stack i.e. a disk can only be moved if it is the uppermost disk on a stack.
3. No disk may be placed on top of a smaller disk.

With three disks, the puzzle can be solved in seven moves. The minimum number of moves required to solve a Tower of Hanoi puzzle is2n - 1, where n is the number of disks. (c) Wikipedia.

SmallY's puzzle is very similar to the famous Tower of Hanoi. In the Tower of Hanoi puzzle you need to solve a puzzle in minimum number of moves, in SmallY's puzzle each move costs some money and you need to solve the same puzzle but for minimal cost. At the beginning of SmallY's puzzle all n disks are on the first rod. Moving a disk from rod i to rod j (1 ≤ i, j ≤ 3) costs tij units of money. The goal of the puzzle is to move all the disks to the third rod.

In the problem you are given matrix t and an integer n. You need to count the minimal cost of solving SmallY's puzzle, consisting of ndisks.

Input

Each of the first three lines contains three integers — matrix t. The j-th integer in the i-th line is tij (1 ≤ tij ≤ 10000; i ≠ j). The following line contains a single integer n (1 ≤ n ≤ 40) — the number of disks.

It is guaranteed that for all i (1 ≤ i ≤ 3), tii = 0.

Output

Print a single integer — the minimum cost of solving SmallY's puzzle.

Sample test(s)

input

0 1 1
1 0 1
1 1 0
3

output

7

input

0 2 2
1 0 100
1 2 0
3

output

19

input

0 2 1
1 0 100
1 2 0
5

output

87

题意:（转）

在原来的汉诺塔基础上,给每次移动加上一个权值,Tij表示从rodi移动到rodj的花费。

求把n个盘子从rod1移动到rod3上的最少花费。

PS:

dp[n][i][j]：表示把 n 个盘子从 rodi 移动到 rodj 上的最少花费。

决策有两个：

1. 将rod1上的n-1个盘子移动到rod2上,将rod1上的最后一个移动到rod3上,再将rod2上的n-1个盘子移动到rod3上。

2. 将rod1上的n-1个盘子移动到rod3上,将rod1上的最后一个移动到rod2上,将rod3上的n-1个盘子移动回rod1上,将rod2上的最后一个盘子移动到rod3上,最后将rod1上的n-1个盘子移动到rod3上。

dp[0][i][j] = 0, 其他置为无穷大.

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
#define LL long long
int t[4][4];
long long dp[47][4][4];
int main()
{
    for(int i = 1; i <= 3; i++)
    {
        for(int j = 1; j <= 3; j++)
        {
            scanf("%d",&t[i][j]);
        }
    }
    int n;
    LL a, b;
    scanf("%d",&n);
    for(int k = 1; k <= n; k++)
    {
        for(int i = 1; i <= 3; i++)
        {
            for(int j = 1; j <= 3; j++)
            {
                if(i != j)
                {
                    a = dp[k-1][i][6-i-j] + t[i][j] + dp[k-1][6-i-j][j];
                    b = dp[k-1][i][j] + t[i][6-i-j] + dp[k-1][j][i]+t[6-i-j][j] + dp[k-1][i][j];
                    dp[k][i][j] = min(a, b);
                }
            }
        }
    }
    cout<<dp[n][1][3]<<endl;;
}