Admiral（雙向BFS + Hash）

Problem Description

Suppose that you are an admiral of a famous naval troop. Our naval forces have got 21 battleships. There are 6 types of battleships. First, we have got one flagship in which the admiral must be and it is denoted by number 0. Others are denoted by number from 1 to 5, each of them has 2, 3, 4, 5, 6 ships of its kind. So, we have got 21 battleships in total and we must take a giant battle against the enemy. Hence, the correct strategy of how to arrange each type of battleships is very important to us. The shape of the battlefield is like the picture that is shown below. To simplify the problem, we consider all battleships have the same rectangular shape. Fortunately, we have already known the optimal state of battleships. As you can see, the battlefield consists of 6 rows. And we have 6 types of battleship, so the optimal state is that all the battleships denoted by number i are located at the i-th row. Hence, each type of battleship corresponds to different color. You are given the initial state of battlefield as input. You can change the state of battlefield by changing the position of flagship with adjacent battleship. Two battleships are considered adjacent if and only if they are not in the same row and share parts of their edges. For example, if we denote the cell which is at i-th row and j-th position from the left as (i,j), then the cell (2,1) is adjacent to the cells (1,0), (1,1), (3,1), (3,2). Your task is to change the position of the battleships minimum times so as to reach the optimal state. Note: All the coordinates are 0-base indexed.

Input

The first line of input contains an integer T (1 <= T <= 10), the number of test cases.  Each test case consists of 6 lines. The i-th line of each test case contains i integers, denoting the type of battleships at i-th row of battlefield, from left to right.

Output

For each test case, if you can’t reach the goal in no more than 20 moves, you must output 「too difficult」 in one line. Otherwise, you must output the answer in one line.

SampleInput

1
1
2 0
2 1 2
3 3 3 3
4 4 4 4 4
5 5 5 5 5 5

SampleOutput

3

題意就是給你一個6*6的塔，上下兩個相鄰的單位能夠進行交換，問最少進行幾回交換，能夠獲得
0
1 1
2 2 2
3 3 3 3
……………………
這種狀態，開始思路是用A*作，結果A*不是很熟練，沒搞出來，寫了個直接搜索炸了，而後我也是看了一下網上博客，使用雙向搜索就好了。
思路就是從末尾開始往前搜索10步，從開始狀態日後搜索10步，分別狀態壓縮一下存在map中，而後就看有沒有兩種相同的狀態，不然就輸出太難了。
代碼：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 int fx[4][2] = {1,0,1,1,-1,-1,-1,0};    //左下，右下，左上，右上
 5 
 6 struct node{
 7     ll p[6][6];
 8     int r,c;
 9     int flag;
10     int step;
11 
12     node(){}
13     node(int _r,int _c,int _flag,int _step):r(_r),c(_c),flag(_flag),step(_step){}
14 };
15 
16 queue<node>q;
17 map<ll,ll>p[2];　　//分別存儲兩個方向的bfs狀態
18 
19 ll _hash(node a){　　//用hash壓縮路徑狀態
20     ll res = 0;
21     for(int i = 0; i < 6; i++){
22         for(int j = 0; j <= i; j++){
23             res = res*6 + a.p[i][j];
24         }
25     }
26     return res;
27 }
28 
29 int bfs(node &s,node &e){
30     while(!q.empty()){
31         q.pop();
32     }
33     p[0].clear();
34     p[1].clear();
35     q.push(s);
36     q.push(e);
37     p[s.flag][_hash(s)] = 0;　　//必需要標記一下，由於後面會用到count函數查詢是否存在
38     p[e.flag][_hash(e)] = 0;
39     while(!q.empty()){
40         node now = q.front();
41         q.pop();
42         ll sta = _hash(now);
43         if(p[!now.flag].count(sta)){
44             int num = p[!now.flag][sta] +  now.step;
45             if(num <= 20)
46                 return num;
47             else
48                 continue;
49         }
50 
51         if(now.step >= 10)　　//處理10步便可
52             continue;
53         for(int i = 0; i < 4; i++){
54             node nxt = now;
55             nxt.step++;
56             nxt.r += fx[i][0];
57             nxt.c += fx[i][1];
58             if(nxt.r < 0 || nxt.r > 6 || nxt.c < 0 || nxt.c > nxt.r)
59                 continue;
60             swap(nxt.p[now.r][now.c],nxt.p[nxt.r][nxt.c]);
61             if(p[nxt.flag].count(_hash(nxt)) == 0)
62                 p[nxt.flag][_hash(nxt)] = nxt.step;
63             q.push(nxt);
64         }
65     }
66     return -1;
67 }
68 
69 int main(){
70     int t;
71     cin>>t;
72     node s, e;
73     while(t--){
74         for(int i = 0; i < 6; i++){
75             for(int j = 0; j <= i; j++){
76                 cin>>s.p[i][j];
77                 if(s.p[i][j] == 0)
78                     s.r = i, s.c = j;
79                 e.p[i][j] = i;
80             }
81         }
82         s.flag = 0;
83         s.step = 0;
84         e = node(0,0,1,0);
85         int ans = bfs(s,e);
86         if(ans >= 0 && ans <= 20)
87             cout << ans << endl;
88         else
89             cout << "too difficult" << endl;
90     }
91     return 0;
92 }