Problem 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).
解题思路：最简单的dfs和bfs，个人还是建议用dfs，时间短，而且还好写；
具体的上代码；
#include <iostream>
#include <cstring>
using namespace std;
char map[30][30];
bool vis[30][30];
int qx[900],qy[900];//bfs
int dir[4][2]= {1,0,-1,0,0,1,0,-1};
int n,m,ans;
/*void bfs(int x, int y)
{
int l=0,r=0;
qx[r]=x,qy[r++]=y;
vis[x][y]=1;
ans++;
int nx,ny;
while(l<r)
{
int curx=qx[l],cury=qy[l++];
for(int i=0; i<4; i++)
{
nx=curx+dir[i][0];
ny=cury+dir[i][1];
if(nx>=0&&nx<n && ny>=0&&ny<m && !vis[nx][ny] && map[nx][ny]!='#')
{
vis[nx][ny]=1;
ans++;
qx[r]=nx,qy[r++]=ny;
}
}
}
}*/
void dfs(int x, int y)
{
ans++;
vis[x][y]=1;//可以不用
map[x][y]='#';
for(int i=0; i<4; i++)
{
int nx=x+dir[i][0];
int ny=y+dir[i][1];
if(nx>=0&&nx<n && ny>=0&&ny<m &&!vis[nx][ny]&& map[nx][ny]!='#')
{
dfs(nx, ny);
}
}
//cout<<ans<<endl;
}
int main()
{
while(cin>>m>>n, m,n)
{
memset(vis, 0, sizeof(vis));
ans=0;
int sx,sy;
for(int i=0; i<n; i++)
{
for(int j=0; j<m; j++)
{
cin>>map[i][j];
if(map[i][j] == '@')
sx=i,sy=j;
}
}
//bfs(sx, sy);
dfs(sx, sy);
cout<<ans<<endl;
}
return 0;
}
/*
6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0
*/