杭电1312 red and black

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14490    Accepted Submission(s): 8965

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

 

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<span style="font-size:32px;">#include<stdio.h>#include<string.h>char map[21][21];int via[21][21],m,n,ans;void dfs(int x,int y){if(x<0||y<0||x>=m||y>=n||via[x][y]||map[x][y]=='#')return ;via[x][y]=1;ans++;dfs(x,y+1);dfs(x,y-1);dfs(x+1,y);dfs(x-1,y);}int main(){int i,j,sx,sy;while(scanf("%d %d",&n,&m),m||n){   ans=0;memset(via,0,sizeof(via));memset(map,0,sizeof(map));for(i=0;i<m;i++){scanf("%s",&map[i]);for(j=0;j<n;j++){if(map[i][j]=='@')sx=i,sy=j;}}dfs(sx,sy);printf("%d\n",ans);}return 0;}</span>