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
45
59
6
13
思路：
先找到，然后用bfs进行搜索，查找此点附近的四个点，如果是'.'并且之前没有走过这个点则把此点加入队列并且标记为'走过的点'，同时可以走的点计数加1。
源代码：
1 #include <iostream>
2 #include<queue>
3 #include <stdio.h>
4 #include<string.h>
5 char s[25][25]; 6 int visit[25][25]; 7 using namespace std; 8 int main() 9 { 10 int h,w; 11 int x,y,a,b,i,j; 12 int sum=0; 13 while(scanf("%d %d",&w,&h)) 14 { 15 if(w==0&&h==0) break; 16 memset(visit,0,sizeof(visit)); 17 sum=0; 18 for(i=0;i<h;i++) 19 { 20 scanf("%s",s[i]); 21
22 for(j=0;j<w;j++) 23 { 24
25 if(s[i][j]=='@') 26 {x=i; 27 y=j; 28 visit[i][j]=1; 29
30 } 31 if(s[i][j]=='#') 32 visit[i][j]=1; 33 } 34 } 35 int ok; 36 queue<int>q; 37 q.push(x); 38 q.push(y); 39 while(!q.empty()) 40 {ok=1; 41 a=q.front(); 42 q.pop(); 43 b=q.front(); 44 q.pop(); 45 if(a-1>=0&&s[a-1][b]=='.'&&visit[a-1][b]==0) 46 { 47 sum++; 48 visit[a-1][b]=1; 49 q.push(a-1); 50 q.push(b); 51
52 } 53 if(a+1<h&&s[a+1][b]=='.'&&visit[a+1][b]==0) 54 { 55 sum++; 56 visit[a+1][b]=1; 57 q.push(a+1); 58 q.push(b); 59
60 } 61 if(b-1>=0&&s[a][b-1]=='.'&&visit[a][b-1]==0) 62 { 63 sum++; 64 visit[a][b-1]=1; 65 q.push(a); 66 q.push(b-1); 67
68 } 69 if(b+1<w&&s[a][b+1]=='.'&&visit[a][b+1]==0) 70 { 71 sum++; 72 visit[a][b+1]=1; 73 q.push(a); 74 q.push(b+1); 75
76 } 77 for(i=0;i<h;i++) 78 { 79 for(j=0;j<w;j++) 80 { 81
82 } 83 } 84 } 85 printf("%d\n",sum+1); 86
87 } 88 return 0; 89 }