red and black(BFS)

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 40685		Accepted: 22079

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) 
The end of the input is indicated by a line consisting of two zeros. 

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

Japan 2004 Domestic

 

题解：简单的BFS

AC代码

#include<stdio.h>
#include<iostream>
#include<stdio.h>
#include<string>
#include<cmath>
#include<algorithm>
using namespace std;

char a[25][25];
int w, h;
int ant;

void dfs(int x, int y)
{
    if(a[x][y] != '#' && x >= 0 && x < h && y >= 0 && y < w)
    {
        ant++;
        a[x][y] = '#';
        dfs(x+1, y);
        dfs(x-1, y);
        dfs(x, y+1);
        dfs(x, y-1);
    }
}

int main()
{
    while(scanf("%d%d", &w, &h) != EOF)
    {
        ant = 0;
        if(w == 0 && h == 0)
            break;
        for(int i = 0; i < h; i++)
        {
            for(int j = 0; j < w; j++)
            {
                cin >> a[i][j];
            }
        }
        for(int i = 0; i < h; i++)
            for(int j = 0; j < w; j++)
                if(a[i][j] == '@')
                    dfs(i, j);
        cout << ant << endl;
    }
    return 0;
}