Red and Black

Problem 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)

 

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

4559613

题解：深搜。

#include <iostream>#include <cstdio>#include <cstring>#include <queue>using namespace std;char map[50][50];int d[4][2] = {{0,-1},{0,1},{-1,0},{1,0}};int n,m;int sx,sy;int ans;void dfs(int x,int y){if(x < 0 || y < 0 || x >= n || y >= m || map[x][y] == '#'){return;}ans++;map[x][y] = '#';for(int i = 0;i < 4;i++){dfs(x + d[i][0],y + d[i][1]);}}int main(){while(scanf("%d%d",&m,&n) != EOF && (m + n) != 0){getchar();for(int i = 0;i < n;i++){for(int j = 0;j < m;j++){scanf("%c",&map[i][j]);if(map[i][j] == '@'){sx = i;sy = j;}}getchar();}ans = 0;dfs(sx,sy);printf("%d\n",ans);}return 0;}