HDU 1312--Red and Black【DFS】

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11503    Accepted Submission(s): 7158

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

 

#include <cstdio>#include <cstring>int n,m;char map[100][100];int dir[][2]={0,1,0,-1,1,0,-1,0};int ans;void dfs(int x,int y){    map[x][y]='#';    for(int i=0;i<4;++i){        int fx=x+dir[i][0];        int fy=y+dir[i][1];        if(fx>=0 && fx<m && fy>=0 && fy<n && map[fx][fy]=='.'){            dfs(fx,fy);            ans++;        }    }}int main (){    int i,j,sx,sy;    while(scanf("%d%d",&n,&m),n,m){        for(i=0;i<m;++i){            scanf("%s",map[i]);            for(j=0;j<n;++j){                if(map[i][j]=='@'){                    sx=i,sy=j;                }            }        }    ans=0;    dfs(sx,sy);    printf("%d\n",ans+1);    }    return 0;}