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)
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
45
59
6
13
深搜实现方案：
1 #include<stdio.h> 2 #include<string.h> 3 #include<algorithm> 4 #include<iostream> 5 using namespace std; 6 char map[110][110]; 7 int vis[110][100]; 8 int dir[8][2]={{0,1},{0,-1},{1,0},{-1,0}}; 9 int n,m,num; 10 void DFS(int x,int y) 11 { 12 int a,b,i; 13 vis[x][y]=1; 14 num++; 15 for(i=0;i<8;i++) 16 { 17 a=x+dir[i][0]; 18 b=y+dir[i][1]; 19 if(a>=0&&a<m&&b>=0&&b<n&&vis[a][b]==0&&map[a][b]=='.') 20 { 21 DFS(a,b); 22 } 23 } 24 } 25 int main() 26 { 27 int i,j,x,y; 28 while(scanf("%d%d",&n,&m)!=EOF) 29 { 30 getchar(); 31 if(m==0&&n==0) 32 break; 33 memset(map,0,sizeof(map)); 34 memset(vis,0,sizeof(vis)); 35 for(i=0; i<m; i++) 36 { 37 scanf("%s",map[i]); 38 } 39 for(i=0; i<m; i++) 40 { 41 for(j=0; j<n; j++) 42 { 43 if(map[i][j]=='@')///只有一个人 44 { 45 x=i; 46 y=j; 47 } 48 } 49 } 50 num=0; 51 DFS(x,y); 52 printf("%d ",num); 53 } 54 return 0; 55 }