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Red and Black

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
[email protected] - 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

最簡單常規的深度優先搜索（DFS）題，與POJ2386這個題很類似。不再詳細解釋，直接上代碼：

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
char maze[25][25];
bool flag[25][25];
int n,m,dir[4][2] = { {1,0},{-1,0},{0,1},{0,-1} },ans;
void dfs(int x, int y)
{
    if (x<=0||x > n ||y<=0|| y > m|| maze[x][y] == ‘#‘)
        return;
    if (flag[x][y])//這個位置未被標記，之前沒被走過
    {
        ans++;
        flag[x][y] = false;
    }
    maze[x][y] = ‘#‘;//對當前位置置換成#，避免下次對這個位置重復DFS 
    for (int i = 0; i < 4; i++)//向四個方向DFS
    {
        int nx = x + dir[i][0], ny = y + dir[i][1];
        dfs(nx, ny);
    }
    return ;
}
int main()
{
    while (cin >> m >> n,(m||n))
    {
        memset(flag, true, sizeof(flag));
        int sx, sy;
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= m; j++)
            {
                cin >> maze[i][j];
                if (maze[i][j] == ‘@‘)
                {
                    sx = i;
                    sy = j;
                }
            }
        ans = 0;
        dfs(sx, sy);
        cout << ans << endl;
    }
    return 0;
}

POJ 1979 Red and Black（簡單DFS）