Red and Black
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 60000/30000K (Java/Other)
Total Submission(s) : 20 Accepted Submission(s) : 17
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)
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
PKU

 ```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 ``` ```#include #include #include #include #define Max 52   using namespace std;   int data[Max][Max]; int vis[Max][Max]; int a,b,count=0; int sum=0,tmp=0; struct mod{ int e,w,n,s; }pas[Max][Max]; void sign(){ int i,j; for(i=1;i<=a;i++) for(j=1;j<=b;j++){ if(!(data[i][j]&1)) pas[i][j].w=1; if(!(data[i][j]&4)) pas[i][j].e=1; if(!(data[i][j]&2)) pas[i][j].n=1; if(!(data[i][j]&8)) pas[i][j].s=1; } } void dfs(int x,int y){ if(!x||!y||x==a+1||y==b+1||vis[x][y]) return ; //int der[4][2] = {{0,1},{0,-1},{-1,0},{1,0}}; vis[x][y]=1; tmp++; if(pas[x][y].w) dfs(x,y-1); if(pas[x][y].e) dfs(x,y+1); if(pas[x][y].n) dfs(x-1,y); if(pas[x][y].s) dfs(x+1,y); } int main(){ int i,j; cin>>a>>b; for(i=1;i<=a;i++) for(j=1;j<=b;j++) scanf("%d",&data[i][j]); sign(); for(i=1;i<=a;i++) for(j=1;j<=b;j++){ //printf("aa"); if(vis[i][j]) continue; //printf("bb"); count++; dfs(i,j); if(tmp>sum) sum=tmp; tmp=0; } cout<