Curling 2.0（DFS简单题）

题目链接：

https://vjudge.net/problem/POJ-3009

题目描述：

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

The input is a sequence of datasets. The end of the input is indicated by a line containing two zeros separated by a space. The number of datasets never exceeds 100.

Each dataset is formatted as follows.

the width(=w) and the height(=h) of the board 
First row of the board 
... 
h-th row of the board

The width and the height of the board satisfy: 2 <= w <= 20, 1 <= h <= 20.

Each line consists of w decimal numbers delimited by a space. The number describes the status of the corresponding square.

0	vacant square
1	block
2	start position
3	goal position

The dataset for Fig. D-1 is as follows:

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

Output

For each dataset, print a line having a decimal integer indicating the minimum number of moves along a route from the start to the goal. If there are no such routes, print -1 instead. Each line should not have any character other than this number.

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

题意描述
给出棋盘，从起始位置到目标位置，如果能够到达，输出最短步数，不能到达输出-1
解题思路：
用DFS进行搜索，搜索时注意先判断是否可以击打，也就是该方向上至少存在一个空地，可以，则模拟向该方向行进，先判断是否走到目的地，如果走到，直接返回，否则看是否在走出边界
之前碰到一块砖，碰到则可以向砖之前的这一块空地上继续向下搜索。四个方向，挨个搜索即可。
代码实现：

#include<stdio.h>
#include<string.h>
int map[][];
int r,c,ex,ey,ans,sx,sy;

void dfs(int x,int y,int s);

int main()
{
    int i,j;
    while(scanf("%d%d",&c,&r), c+r != )
    {
        memset(map,,sizeof(map));
        for(i=;i<=r;++i)
            for(j=;j<=c;++j){
                scanf("%d",&map[i][j]);
                if(map[i][j]==){
                    sx=i;sy=j;
                    map[i][j]=;
                }
                if(map[i][j]==){
                    ex=i;ey=j;
                    map[i][j]=;
                }
            }

        ans=;
        dfs(sx,sy,);

        if(ans==)
        printf("-1\n");
        else
        printf("%d\n",ans);
    }
    return ;
}
void dfs(int x,int y,int s)
{
    if(s > )
    return;

    int tx,ty;
    //up
    tx=x-;
    ty=y;
    if(!map[tx][ty])
    {
        for(;tx>=;tx--)
        {
            if(tx==ex && ty==ey)
            {
                if(s < ans)
                    ans=s;
                return;
            }
            if(map[tx][ty])
                break;
        }
        if(map[tx][ty])
        {
            map[tx][ty]=;
            dfs(tx+,ty,s+);

            map[tx][ty]=;
        }
    }
    //right
    tx=x;
    ty=y+;
    if(!map[tx][ty])
    {
        for(;ty<=c;ty++)
        {
            if(tx==ex && ty==ey)
            {
                if(s < ans)
                    ans=s;
                return;
            }
            if(map[tx][ty])
                break;
        }
        if(map[tx][ty])
        {
            map[tx][ty]=;
            dfs(tx,ty-,s+);

            map[tx][ty]=;
        }
    }
    //down
    tx=x+;
    ty=y;
    if(!map[tx][ty])
    {
        for(;tx<=r;tx++)
        {
            if(tx==ex && ty==ey)
            {
                if(s < ans)
                    ans=s;
                return;
            }
            if(map[tx][ty])
                break;
        }
        if(map[tx][ty])
        {
            map[tx][ty]=;
            dfs(tx-,ty,s+);

            map[tx][ty]=;
        }
    }
    //left
    tx=x;
    ty=y-;
    if(!map[tx][ty])
    {
        for(;ty>=;ty--)
        {
            if(tx==ex && ty==ey)
            {
                if(s < ans)
                    ans=s;
                return;
            }
            if(map[tx][ty])
                break;
        }
        if(map[tx][ty])
        {
            map[tx][ty]=;
            dfs(tx,ty+,s+);

            map[tx][ty]=;
        }
    }
}

易错分析：

注意起步时步数为1

注意超出边界和走到目的地的情况与碰到砖块处理