toj 2815 Searching Problem

2815.   Searching Problem

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Time Limit: 1.0 Seconds    Memory Limit: 65536K
Total Runs: 421    Accepted Runs: 177

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It is said that the predecessors had hidden a lot of expensive treasures just before the death of their city hundreds of years ago. People's searching for it has never stopped. The only problem is that they can't tell the exact position of the treasure. What's more, there exists great dangerousness in searching. So although the treasure has been their place for countries, it's still safe.

You are one of the people who want to make a fortune. However, you're a very lucky man. You happen to get a secret map, and you know the map can lead you to the treasure directly.

The map is a maze. There are five symbols on the map. They are 'S', 'D', '$', '1' and '0'. 'S' is the point you enter into the maze. '$' stand for the treasure in the maze. '0' and 'D' are the points you can walk along, while '1' is walls that you can't go through. You shall start at the 'S' point, and pick up the treasure in the maze along the '0' and 'D' points, then come out along the same way.

It seems easy, and you'll become a wealthy man. But there is still a problem you have to solve. On every point marked as a 'D' there is a dog to protect the treasure. It's absurd that the dogs can live for countries without food! Just forget about it. So you decide to bring a gun with you. But how many bullets it will take? It's up to you to calculate here. Assume that you're a good shooter, and need only one bullet to shoot one dog.

Input

The first line of each test case is two integer R and C (1 ≤ R, C ≤ 50). R is the row number of the maze, and C is the column number of the maze. The following R lines will contain C characters for each. The R * C characters includes exactly one 'S' and one '$', the remains are '1', '0' and 'D'. It guarantees that the 'S' point is always on the edge of the maze.

Two zeros indicate the end of the input.

Output

For each case, just print a line with the number K, which is the least number of bullets you need.

Sample Input

4 5
00$00
00100
D00D0
S0001
4 5
00$00
01100
D00D0
S0001
4 5
DDDDD
DD$DD
D0D0D
SDD0D
0 0

Sample Output

0
1
2

Problem Setter: Venice

Source: TJU Programming Contest 2007

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#include  < iostream >
#include  < queue >
#define  MAX 54
using   namespace  std;
int  mark[MAX][MAX],a[ 4 ][ 2 ] = {{ - 1 , 0 },{ 1 , 0 },{ 0 , 1 },{ 0 , - 1 }};
char  map[MAX][MAX];
typedef  struct  node
{
     int  x;
     int  y;
     int  step;
}Point;
queue < Point > Q;
Point start,p,temp;
int  r,c,MM;
void  Init()
{
     int  i,j,k;
     for (i = 0 ;i < r;i ++ )
         for (j = 0 ;j < c;j ++ )
        {
            cin >> map[i][j];
             if (map[i][j] == ' S ' )
            {
                start.x = i;
                start.y = j;
                start.step = 0 ;
            }
            mark[i][j] =- 1 ;
        }
}
bool  Bound( int  x, int  y)
{
     if (x >= 0 && y >= 0 && x < r && y < c)
         return   true ;
     else
         return   false ;
}
void  BFS()
{
     int  k,tx,ty;
     while ( ! Q.empty())
        Q.pop();
    Q.push(start);
    mark[start.x][start.y] = 0 ;
    MM =- 1 ;
     while ( ! Q.empty())
    {
        p = Q.front();
        Q.pop();
         if (map[p.x][p.y] == ' $ ' )
        {
             if (p.step < MM || MM ==- 1 )
            {
                MM = p.step;
            }
        }
         for (k = 0 ;k < 4 ;k ++ )
        {
            tx = p.x + a[k][ 0 ];
            ty = p.y + a[k][ 1 ];
             if ( ! Bound(tx,ty) || map[tx][ty] == ' 1 ' )
                 continue ;
             if (map[tx][ty] == ' 0 ' || map[tx][ty] == ' $ ' )
            {
                 if (mark[tx][ty] ==- 1 || p.step < mark[tx][ty])
                {
                    temp.x = tx;
                    temp.y = ty;
                    temp.step = p.step;
                    mark[tx][ty] = p.step;
                    Q.push(temp);
                }
                 continue ;
            }
             if (map[tx][ty] == ' D ' )
            {
                temp.step = p.step + 1 ;
                 if (mark[tx][ty] ==- 1 || temp.step < mark[tx][ty])
                {
                    temp.x = tx;
                    temp.y = ty;
                     // temp.step=p.step;
                    mark[tx][ty] = temp.step;
                    Q.push(temp);
                }
                 continue ;
            }
        }
    }
    printf( " %d\n " ,MM);
}
int  main()
{
     while (scanf( " %d%d " , & r, & c) != EOF)
    {
         if (r == 0 && c == 0 )
             break ;
        Init();
        BFS();
    }
     return   0 ;
}

转载于:https://www.cnblogs.com/forever4444/archive/2009/05/14/1456965.html