HPU DFS + BFS 专项练习A

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

题意：有一个M*N的地板，分别有红地板‘#’和黑地板‘.’组成，一个人站在一块黑地板‘@’上并且只能走黑地板，问能走多少块黑地板；

思路：简单DFS:

下面附上模板：

#include<bits/stdc++.h>using namespace std;char mapp[105][105];int vis[25][25];int fx[4]={0,0,-1,1},fy[4]={1,-1,0,0};int W,H;int ans=0;void dfs(int x, int y){ans++;vis[x][y]=1;for(int i=0;i<4;i++){int nx=x+fx[i];int ny=y+fy[i];if(nx>=0 && ny>=0 && nx<H && ny<W && !vis[nx][ny] && mapp[nx][ny]=='.'){//printf("%d %d\n",nx,ny);dfs(nx,ny);}}}int main(){while(~scanf("%d %d",&W,&H),W){int sx,sy;ans=0;int flag=0;memset(vis,0,sizeof(vis));for(int i=0;i<H;i++)scanf("%s",mapp[i]);for(int i=0;i<H;i++){for(int j=0;j<W;j++)if(!vis[i][j]&&mapp[i][j]=='@'){sy=j;sx=i;flag=1;break;}if(flag) break;}//printf("%d %d\n",sx,sy);dfs(sx,sy);printf("%d\n",ans);}return 0;}