Red and Black

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6865    Accepted Submission(s): 4359

Problem Description

There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles.

Write a program to count the number of black tiles which he can reach by repeating the moves described above. 

 

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

'.' - a black tile 
'#' - a red tile 
'@' - a man on a black tile(appears exactly once in a data set) 

 

Output

For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself). 

 

Sample Input

6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0

 

Sample Output

45 59 6 13

 

思路：确实是简单dfs，初学者一定要多想

 

#include<stdio.h>
#include<string.h>
char str[22][22];
int temp[22][22];
int cnt,h,w;
void dfs(int x,int y)
{
    cnt ++;
    temp[x][y] = 1;
    if(!temp[x+1][y]&& x+1 < w)
        dfs(x+1,y);
    if(!temp[x-1][y] && x-1 > -1)
        dfs(x-1,y);
    if(!temp[x][y+1] && y+1 < h)
        dfs(x,y+1);
    if(!temp[x][y-1] && y-1 > -1)
        dfs(x,y-1);
}

int main()
{
    int i,j,a,b;
    while(~scanf("%d%d",&h,&w)&&h+w)
    {
        cnt = 0;
        for(i = 0;i < w;i ++)
        {
            scanf("%s",str[i]);
            for(j = 0;j < h;j ++)
            {
                if(str[i][j] == '#')
                    temp[i][j] = 1;
                if(str[i][j] == '.')
                    temp[i][j] = 0;
                if(str[i][j] == '@')
                {
                    a = i;
                    b = j;
                }
            }
        }
        dfs(a,b);
        printf("%d
",cnt);
    }
    return 0;
}