POJ 1979 Red and Black （DFS）

/*简单的dfs题，注意map数组清零

注意题目中地图输入的行列顺序，陷阱！*/

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

4559613

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int n,m,mmax;char map[40][40];int a[4]={0,1,0,-1},b[4]={1,0,-1,0};int cns;void dfs(int si,int sj){if(si<0||sj<0||si>=n||sj>=m) return;for(int i=0;i<4;i++){if(map[si+a[i]][sj+b[i]]=='.'){cns++;map[si+a[i]][sj+b[i]]='#';dfs(si+a[i],sj+b[i]);}}}int main(){int i,j,si,sj;char ch;while(scanf("%d%d",&m,&n)!=EOF){if(n==0&&m==0) break;memset(map,0,sizeof(map));for(i=0;i<n;i++){for(j=0;j<m;j++){cin>>ch;map[i][j]=ch;if(ch=='@'){si=i;sj=j;}}}map[si][sj]='#';cns=1;dfs(si,sj);printf("%d\n",cns);}return 0;}