【HDU1312】A

题目链接

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

题目意思就是从  @  开始可以走的  ‘.’  最多有多少个

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;const int MAX = 22;typedef long long LL;char s[MAX][MAX];int vis[MAX][MAX],w,h,ans;int fx[4]={0,0,1,-1};int fy[4]={1,-1,0,0};void dfs(int x,int y){vis[x][y]=1;ans++;for(int i=0;i<4;i++)   //可以走的四个方向{int xx=x+fx[i],yy=fy[i]+y;if(xx>=0&&yy>=0&&xx<h&&yy<w&&!vis[xx][yy]&&s[xx][yy]=='.')dfs(xx,yy);}}int main(){while(~scanf("%d%d",&w,&h),h){int x,y;memset(vis,0,sizeof(vis));     //初始化为0for(int i=0;i<h;i++)scanf("%s",s[i]);for(int i=0;i<h;i++)for(int j=0;j<w;j++){if(s[i][j]=='@')    //找到@的位置x=i,y=j;}ans=0;dfs(x,y);printf("%d\n",ans);}return 0;}