Curling 2.0

On Planet MM-21, after their Olympic games this year, curling is getting popular. But the rules are somewhat different from ours. The game is played on an ice game board on which a square mesh is marked. They use only a single stone. The purpose of the game is to lead the stone from the start to the goal with the minimum number of moves.

Fig. 1 shows an example of a game board. Some squares may be occupied with blocks. There are two special squares namely the start and the goal, which are not occupied with blocks. (These two squares are distinct.) Once the stone begins to move, it will proceed until it hits a block. In order to bring the stone to the goal, you may have to stop the stone by hitting it against a block, and throw again

The movement of the stone obeys the following rules:

• At the beginning, the stone stands still at the start square.
• The movements of the stone are restricted to x and y directions. Diagonal moves are prohibited.
• When the stone stands still, you can make it moving by throwing it. You may throw it to any direction unless it is blocked immediately(Fig. 2(a)).
• Once thrown, the stone keeps moving to the same direction until one of the following occurs:
  • The stone hits a block (Fig. 2(b), (c)).
    • The stone stops at the square next to the block it hit.
    • The block disappears.
  • The stone gets out of the board.
    • The game ends in failure.
  • The stone reaches the goal square.
    • The stone stops there and the game ends in success.
• You cannot throw the stone more than 10 times in a game. If the stone does not reach the goal in 10 moves, the game ends in failure.

Under the rules, we would like to know whether the stone at the start can reach the goal and, if yes, the minimum number of moves required.

With the initial configuration shown in Fig. 1, 4 moves are required to bring the stone from the start to the goal. The route is shown in Fig. 3(a). Notice when the stone reaches the goal, the board configuration has changed as in Fig. 3(b).

Fig. 3: The solution for Fig. D-1 and the final board configuration

Input

Output

Sample Input

2 1
3 2
6 6
1 0 0 2 1 0
1 1 0 0 0 0
0 0 0 0 0 3
0 0 0 0 0 0
1 0 0 0 0 1
0 1 1 1 1 1
6 1
1 1 2 1 1 3
6 1
1 0 2 1 1 3
12 1
2 0 1 1 1 1 1 1 1 1 1 3
13 1
2 0 1 1 1 1 1 1 1 1 1 1 3
0 0

Sample Output

1
4
-1
4
10
-1

翻译：一个n*m的矩形格子，其中有包含一个数字“2”和一个数字“3”，剩下的格子由“0”和“1”组成，目的是计算从“2”走到“3”的最短步数，“1”代表障碍物，“0”代表可以通行。“2”可以往周围四个方向走，如果往某一个方向走，
那么停下来的条件是，当这个方向上存在障碍物“1”，且会停在这个障碍物的前一个格子，并会击碎这个障碍物;如果选择的方向上没有障碍物“1”也没有终点“3”，那么就会一直运动下去，直到出界。如果遇到“3”，
那么就算一种到达终点的方法。每选择往一个方向运动，就算一步。如果不能到达或者到达的步数大于10,那么就算失败，输出-1.

思路：DFS搜索整个解答树，外加模拟

 1 #include <cstdio>
 2 #include <fstream>
 3 #include <algorithm>
 4 #include <cmath>
 5 #include <deque>
 6 #include <vector>
 7 #include <queue>
 8 #include <string>
 9 #include <cstring>
10 #include <map>
11 #include <stack>
12 #include <set>
13 #include <sstream>
14 #include <iostream>
15 #define mod 1000000007
16 #define eps 1e-6
17 #define ll long long
18 #define INF 0x3f3f3f3f
19 using namespace std;
20 
21 int m,n;
22 int gc[25][25];
23 int fx[4]={-1,1,0,0},fy[4]={0,0,-1,1};
24 int ans;
25 int pd(int i,int j)
26 {
27     if(gc[i][j]==0||gc[i][j]==2)
28     {
29         return 1;
30     }
31     else if(gc[i][j]==-1||gc[i][j]==1)
32     {
33         return 2;
34     }
35     else
36     {
37         return 3;
38     }
39 }
40 void dfs(int x,int y,int t)
41 {
42     if(gc[x][y]==3||t>10)
43     {
44         ans=(t<ans?t:ans);
45     }
46     else
47     {
48         for(int i=0;i<4;i++)
49         {
50             int tx=x,ty=y;
51             if(pd(tx+fx[i],y+fy[i])!=2)
52             {
53                 while(pd(tx+fx[i],ty+fy[i])==1)
54                 {
55                     tx+=fx[i];
56                     ty+=fy[i];
57                 }
58                 if(gc[tx+fx[i]][ty+fy[i]]==1)
59                 {
60                     gc[tx+fx[i]][ty+fy[i]]=0;
61                     t++;
62                     dfs(tx,ty,t);
63                     t--;
64                     gc[tx+fx[i]][ty+fy[i]]=1;
65                 }
66                 else if(gc[tx+fx[i]][ty+fy[i]]==3)
67                 {
68                     t++;
69                     dfs(tx+fx[i],ty+fy[i],t);
70                     t--;
71                 }
72             }
73         }
74     }
75 }
76 int main()
77 {
78     while(cin>>m>>n,m!=0,n!=0)
79     {
80         int x,y;
81         memset(gc,-1,sizeof(gc));
82         for(int i=1;i<=n;i++)
83         {
84             for(int j=1;j<=m;j++)
85             {
86                 cin>>gc[i][j];
87                 if(gc[i][j]==2)
88                 {
89                     x=i;
90                     y=j;
91                 }
92             }
93         }
94         ans=10000000;
95         dfs(x,y,0);
96         printf("%d\n",ans>10?-1:ans);
97     }
98 }

Curling 2.0

原文：https://www.cnblogs.com/mzchuan/p/11180385.html