Red and Black（杭电oj1312）（BFS）

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10122    Accepted Submission(s): 6313

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

 

Source

Asia 2004, Ehime (Japan), Japan Domestic

#include<stdio.h>char s[25][25];int a[4]={0,0,1,-1};int b[4]={1,-1,0,0};int m,n,sum;void bfs(int x,int y){int k,v,t;    s[x][y]='#';for(k=0;k<4;k++){v=x+a[k];t=y+b[k];if(s[v][t]=='.'&&v>=1&&v<=n&&t>=0&&t<m){bfs(v,t);sum++;}}}int main(){int i,j;while(scanf("%d%d",&m,&n)&&(m+n)){for(i=1;i<=n;i++)scanf("%s",s[i]);for(i=1,sum=1;i<=n;i++){for(j=0;j<m;j++){   if(s[i][j]=='@')   bfs(i,j);    }}printf("%d\n",sum);}return 0;}