HDU1312 Red and Black

標籤：hdu1312

Red and BlackTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9732    Accepted Submission(s): 6060

Problem DescriptionThere 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.
 
InputThe 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)
 
OutputFor 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.#..........#.#######..#.#.....#..#.#.###.#..#.#[email protected]#.#..#.#####.#..#.......#..#########............11 6..#..#..#....#..#..#....#..#..###..#..#..#@...#..#..#....#..#..#..7 7..#.#....#.#..###.###[email protected]###.###..#.#....#.#..0 0

 
Sample Output

4559613

#include <stdio.h>#include <string.h>#define maxn 22int n, m, ans;const int mov[][2] = {1, 0, -1, 0, 0, 1, 0, -1};char map[maxn][maxn];bool check(int x, int y){return x >= 0 && y >= 0 && x < n && y < m && map[x][y] != '#';}void DFS(int x, int y){int a, b, i;for(i = 0; i < 4; ++i){a = x + mov[i][0];b = y + mov[i][1];if(check(a, b)){++ans; map[a][b] = '#';DFS(a, b);}}}int main(){int i, j, x, y;while(scanf("%d%d", &m, &n) == 2 && (m || n)){for(i = 0; i < n; ++i){getchar();for(j = 0; j < m; ++j){map[i][j] = getchar();if(map[i][j] == '@'){x = i; y = j;}}}map[x][y] = '#';ans = 1; DFS(x, y);printf("%d\n", ans);}return 0;}