CF：392B - Tower of Hanoi 记忆化搜索DP

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 11 0 11 1 03

output

7

input

0 2 21 0 1001 2 03

output

19

input

0 2 11 0 1001 2 05

output

87

题意：把n个盘子从第一个柱子放到第三个柱子的最少花费。

思路：

汉诺塔问题的移动方案其实就两种：

第一种：n-1个盘子从1通过3移动到2，最下面的从1移动到3，n-1个盘子从2通过1移动到3.

第二种：n-1个盘子从1通过2移动到3，最下面的从1移动到2，n-1个盘子从3通过2移动到1，最下面的从2移动到3，n-1个盘子从1通过2移动到3.

实际就是最下面盘子的两种移动方法..然后令Dp[n][i][j][k]表示n个盘子从i通过j移动到k的花费，记忆化搜索就行了。

#include <iostream>#include <cstdio>#include <fstream>#include <algorithm>#include <cmath>#include <deque>#include <vector>#include <list>#include <queue>#include <string>#include <cstring>#include <map>#include <stack>#include <set>#define PI acos(-1.0)#define mem(a,b) memset(a,b,sizeof(a))#define sca(a) scanf("%d",&a)#define pri(a) printf("%d\n",a)#define MM 10002#define MN 1005#define INF 168430090using namespace std;typedef long long ll;ll dp[50][4][4][4],a[3][4];int n,i,j;ll solve(int n,int a1,int a2,int a3){    if(!n) return 0;    if(dp[n][a1][a2][a3]>=0) return dp[n][a1][a2][a3];    ll b1=solve(n-1,a1,a3,a2)+a[a1][a3]+solve(n-1,a2,a1,a3);    ll b2=solve(n-1,a1,a2,a3)+a[a1][a2]+solve(n-1,a3,a2,a1)+a[a2][a3]+solve(n-1,a1,a2,a3);    dp[n][a1][a2][a3]=min(b1,b2);    return dp[n][a1][a2][a3];}int main(){    mem(dp,-1);    for(i=1;i<=3;i++)        for(j=1;j<=3;j++)            cin>>a[i][j];    cin>>n;    cout<<solve(n,1,2,3)<<endl;    return 0;}