POJ1979（DFS深度优先搜索）之 Red and Black

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

AC代码如下：

#include <iostream>#include <cstdio>#include <cstring>using namespace std; const int maxn=25;char a[maxn][maxn];int d[maxn][maxn];int count;int n,m;void dfs(int row,int col,int num){if(a[row][col]=='#') return;if(row<0 || row>=n || col<0 || col>=m) return;if(d[row][col]!=0) return;d[row][col]=num;for(int i=-1;i<=1;i++)for(int j=-1;j<=1;j++)if(i==0||j==0){if(row+i>=0 && row+i<n && col+j>=0 && col+j<m && a[row+i][col+j]!='#' && a[row+i][col+j]=='.' && d[row+i][col+j] ==0)dfs(row+i,col+j,++count);}}/* run this program using the console pauser or add your own getch, system("pause") or input loop */int main(int argc, char** argv) {while(cin>>m>>n && m && n){for(int i=0;i<n;i++){for(int j=0;j<m;j++){cin>>a[i][j];}}count=1;memset(d,0,sizeof(d));for(int i=0;i<n;i++)for(int j=0;j<m;j++)if(a[i][j]=='@'){dfs(i,j,1);break;}cout<<count<<endl;}return 0;}