POJ 3009 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.

Fig. 1: Example of board (S: start, G: goal)

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.

Fig. 2: Stone movements

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

Source

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <vector>
 8 #include <queue>
 9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int dir[4][2] = {0,-1,-1,0,1,0,0,1};
18 int mp[110][110],w,h,sx,sy,ans,limit;
19 bool can(int x,int y){
20     return (x < h && x >= 0 && y >= 0 && y < w);
21 }
22 bool dfs(int x,int y,int step){
23     if(step >= limit) return false;
24     for(int i = 0; i < 4; i++){
25             int nx = x;
26             int ny = y;
27             bool flag = false;
28             while(can(nx+dir[i][0],ny+dir[i][1]) && mp[nx+dir[i][0]][ny+dir[i][1]] != 1){
29                 nx += dir[i][0];
30                 ny += dir[i][1];
31                 flag = true;
32                 if(mp[nx][ny] == 3) return true;
33             }
34             if(flag && can(nx+dir[i][0],ny+dir[i][1]) && mp[nx+dir[i][0]][ny+dir[i][1]] == 1){
35                 mp[nx+dir[i][0]][ny+dir[i][1]] = 0;
36                 if(dfs(nx,ny,step+1)) return true;
37                 mp[nx+dir[i][0]][ny+dir[i][1]] = 1;
38             }
39     }
40     return false;
41 }
42 int main() {
43     while(scanf("%d %d",&w,&h),w||h){
44         for(int i = 0; i < h; i++){
45             for(int j = 0; j < w; j++){
46                 scanf("%d",mp[i]+j);
47                 if(mp[i][j] == 2){
48                     sx = i;
49                     sy = j;
50                 }
51             }
52         }
53         for(limit = 1; limit <= 10; limit++){
54             if(dfs(sx,sy,0)) break;
55         }
56         limit <= 10?printf("%d\n",limit):puts("-1");
57     }
58     return 0;
59 }

View Code