POJ 1979 Red and Black（简单DFS）

Posted 2020-09-22 Farewell

 

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 
\'@\' - 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

最简单常规的深度优先搜索（DFS）题，与POJ2386这个题很类似。不再详细解释，直接上代码：

#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])//这个位置未被标记，之前没被走过
    {
        ans++;
        flag[x][y] = false;
    }
    maze[x][y] = \'#\';//对当前位置置换成#，避免下次对这个位置重复DFS 
    for (int i = 0; i < 4; i++)//向四个方向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;
}