poj-1979-Red and Black【DFS】

Red and Black

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 30684 Accepted: 16701

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

DFS原理图：

有些小细节要注意：

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;int dx[4]={-1,1,0,0};int dy[4]={0,0,-1,1};int x,y,W,H,cnt;bool vis[50][50];char map[50][50];void f(int a,int b){for(int i=0;i<4;i++){int nx=a+dx[i];int ny=b+dy[i];if(map[nx][ny]=='.'&&!vis[nx][ny]&&nx>=0&&nx<H&&ny>=0&&ny<W){cnt++;vis[nx][ny]=1;f(nx,ny);}}}int main(){while(scanf("%d %d",&W,&H)&&(W||H)){memset(vis,0,sizeof(vis));for(int i=0;i<H;i++)   //注意是 H 行，W 列 {getchar();  // 输入数据每行结束都有一个回车，用 getchar() 吸收 for(int j=0;j<W;j++){scanf("%c",&map[i][j]);if(map[i][j]=='@'){x=i;y=j;vis[x][y]=1;}}}cnt=1;f(x,y);printf("%d\n",cnt);}return 0;}