广度搜索

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;
}