HDU 1312 Red and Black【深搜】

 

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11531    Accepted Submission(s): 7177

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

 

 

Source

 

Asia 2004, Ehime (Japan), Japan Domestic

 

 

 

#include <stdio.h>
#include <string.h>

int n,m,cnt;
char map[30][30];
int to[4][2] = {{1,0},{0,1},{-1,0},{0,-1}};

void dfs(int i,int j)
{
	int k,x,y; 
    cnt++;
    map[i][j] = '#';
    for(k = 0; k<4; k++)
    {
        x = i + to[k][0];
        y = j + to[k][1];
        if(x<n && y<m && x>=0 && y>=0 && map[x][y] == '.')//控制 x，y不讓它越界 
            dfs(x,y);
    }
}

int main()
{
    int i,j,fi,fj;
    while(scanf("%d%d%",&m,&n))//橫向爲 n，縱向爲 m 
    {
    	if(m==0&&n==0)	break;
        for(i = 0; i<n; i++)
        {
            for(j = 0; j<m; j++)
            {
                scanf("%c",&map[i][j]);
                if(map[i][j] == '@')//找到剛開始人所在的位置 
                {
                    fi = i;
                    fj = j;
                }
            }
            getchar();
        }
        cnt = 0;
        dfs(fi,fj);
        printf("%d\n",cnt);
    }

    return 0;
}