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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) 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
Source 
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 #define N 26 6 int n,m; 7 char mp[N][N]; 8 int x,y; 9 int vis[N][N]; 10 int ans; 11 int dirx[]={0,0,-1,1}; 12 int diry[]={-1,1,0,0}; 13 void dfs(int sx,int sy){ 14 for(int i=0;i<4;i++){ 15 int xx=sx+dirx[i]; 16 int yy=sy+diry[i]; 17 if(vis[xx][yy]) continue; 18 if(mp[xx][yy]==‘#‘) continue; 19 if(xx<0 || xx>=n || yy<0 || yy>=m) continue; 20 vis[xx][yy]=1; 21 ans++; 22 dfs(xx,yy); 23 } 24 } 25 int main() 26 { 27 while(scanf("%d%d",&m,&n)==2 && n+m){ 28 for(int i=0;i<n;i++){ 29 scanf("%s",mp[i]); 30 for(int j=0;j<m;j++){ 31 if(mp[i][j]==‘@‘){ 32 x=i;y=j; 33 } 34 } 35 } 36 memset(vis,0,sizeof(vis)); 37 ans=1; 38 vis[x][y]=1; 39 dfs(x,y); 40 printf("%d\n",ans); 41 42 } 43 return 0; 44 }
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原文地址：http://www.cnblogs.com/UniqueColor/p/4767670.html
