Red and Black
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 26944 | Accepted: 14637 |
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.
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.
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
/*题目大意：在一个矩形房间里，里面填满着红砖与黑砖，每次走动只能走黑砖位置，而不能走红砖，
* 可以上下左右四个方向走动，现在初始位置在黑砖位置上，试问从此黑砖位置开始能够到达的砖的数量是多少
*算法分析：从起始位置开始四个方向进行dfs，若遇到红砖位置则将此砖块换为黑砖。换砖的次数即为到达的黑砖数量 */ #include <iostream>
#include <cstdio>
using namespace std; char a[25][25];
int n, m;
int dir[4][2] = {0, 1, 1, 0, 0, -1, -1, 0}; //四个方向数组 void dfs(int x, int y, int &res) {
a[x][y] = '#';
res ++ ;
for (int i = 0; i<4; i++) {
int nx = x + dir[i][0];
int ny = y + dir[i][1];
if (nx>=0 && nx<n && ny>=0 && ny<m && a[nx][ny] == '.')
dfs(nx, ny, res);
}
} int main() { while (cin >> m >> n && (n+m)) { memset(a, 0, sizeof(a));
for (int i = 0; i<n; i++) {
for (int j = 0; j<m; j++) {
cin >> a[i][j];
}
}
int res = 0;
for (int i = 0; i<n; i++) {
for (int j = 0; j<m; j++) {
if (a[i][j] == '@')
dfs(i, j, res);
}
}
cout << res << endl;
}
return 0;
}