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).
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=30;
int n,m,ans;
char map[maxn][maxn];
bool vis[maxn][maxn];
int qx[maxn*maxn],qy[maxn*maxn];
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
void bfs(int x,int y)
{
int l=0,r=0;
qx[r]=x;qy[r]=y;r++;
vis[x][y]=1;
ans++;
while(l<r)
{
int curx=qx[l],cury=qy[l];l++;//当前位置
for(int i=0;i<4;i++)
{
int nx=curx+dir[i][0];
int ny=cury+dir[i][1];
if(nx>=1&&nx<=n&&ny>=1&&ny<=m&&!vis[nx][ny]&&map[nx][ny]!=‘#‘)
{
ans++;
vis[nx][ny]=1;
qx[r]=nx;
qy[r]=ny;
r++;
}
}
}
}
int main()
{
int i,j,sx,sy;
while(scanf("%d%d",&m,&n)==2&&(n||m))
{
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
{
cin>>map[i][j];
if(map[i][j]==‘@‘)
sx=i,sy=j;
}
ans=0;
memset(vis,0,sizeof(vis));
bfs(sx,sy);
cout<<ans<<endl;
}
return 0;
}