HIT-Red and Black DFS

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

45
59
6
13

题目链接： http://acm.hit.edu.cn/hoj/problem/view?id=4000098

用系统栈DFS搜索：#include<stdio.h> #include<memory.h> const int MAXN=20+2; void dfs(int i,int j); char maze[MAXN][MAXN],vis[MAXN][MAXN]; int w,h; int main() { //freopen("in.in","r",stdin); int i,j,k; while(scanf("%d%d",&w,&h)&&(w||h)) { memset(vis,0,sizeof(vis)); for(i=0;i<h;) scanf("%s",maze[i++]); for(i=0,k=0;i<h;i++){ for(j=0;j<w;j++) if(maze[i][j]=='@'){k=1;break;} if(k)break; } dfs(i,j); for(i=0,k=0;i<h;i++) for(j=0;j<w;j++) if(vis[i][j])k++; printf("%d\n",k); } return 0; } void dfs(int i,int j) { if(i<0||i>=h||j<0||j>=w||vis[i][j]||maze[i][j]=='#')return; vis[i][j]=1; dfs(i-1,j); dfs(i,j-1); dfs(i,j+1); dfs(i+1,j); }