TC-SRM391-div2-SortingGame（BFS,STL）

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 };