Problem Statement for SortingGame
Problem Statement
In The Sorting Game, you are given a sequence containing a permutation of the integers between 1 and n, inclusive. In one move, you can take any k consecutive elements of the sequence and reverse their order. The goal of the game is to sort the sequence in ascending order. You are given a int[] board describing the initial sequence. Return the fewest number of moves necessary to finish the game successfully, or -1 if it's impossible.
Definition
Class:
SortingGame
Method:
fewestMoves
Parameters:
int[], int
Returns:
int
Method signature:
int fewestMoves(int[] board, int k)
(be sure your method is public)
Constraints
-
board will contain between 2 and 8 elements, inclusive.
-
Each integer between 1 and the size of board, inclusive, will appear in board exactly once.
-
k will be between 2 and the size of board, inclusive.
Examples
0)
{1,2,3}3
Returns: 0
The sequence is already sorted, so we don't need any moves.
1)
{3,2,1}3
Returns: 1
We can reverse the whole sequence with one move here.
2)
{5,4,3,2,1}2
Returns: 10
This one is more complex.
3)
{3,2,4,1,5}4
Returns: -1
4)
{7,2,1,6,8,4,3,5}4
Returns: 7
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This problem was used for:
Single Round Match 397 Round 1 - Division I, Level One
Single Round Match 397 Round 1 - Division II, Level Two
思路：
因为每组数据最多只有8个数，可以直接进行BFS
在BFS过程中遇到了一个问题：如何记录状态？这是一个最多8维的状态
我想法到了用HASH来解决这个问题，可是构造不出这个HASH函数
实在没办法了，找到了官方题解，竟然发现了一个map的奇技淫巧
map<vector<int>,int>
这TM也行。。。是在下输了。。。。
C++是最好的语言！！！
代码：
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 class SortingGame 6 { 7 public: 8 int fewestMoves(vector <int> board, int k) 9 { 10 queue<vector<int>> q; 11 map<vector<int>,int> vis; 12 q.push(board);vis[board]=0; 13 while(!q.empty()){ 14 cout<<11111<<endl; 15 vector<int> vec=q.front(); 16 q.pop(); 17 int num=vis[vec]; 18 int r=vec.size(); 19 int i=0; 20 for(i=0;i<r-1;i++){ 21 if(vec[i+1]<vec[i]) break; 22 } 23 if(i==r-1) return num; 24 for(int i=0;i+k<=r;i++){ 25 vector<int> tmp=vec; 26 reverse(tmp.begin()+i,tmp.begin()+i+k); 27 if(vis.count(tmp)==0){ 28 q.push(tmp); 29 vis[tmp]=num+1; 30 } 31 } 32 } 33 return -1; 34 } 35 };