Red and Black （简单dfs）

 

题目：https://vjudge.net/problem/POJ-1979

 

 

 

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)
The end of the input is indicated by a line consisting of two zeros.

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 Out

  45

59
6
13


题意：
从@点出发，，问能走多少个. 

思路：简单dfs
 
代码：  

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<set>
#include<algorithm>
#include<map>
#define maxn 200005
using namespace std;
int x[4]={-1,1,0,0};
int y[4]={0,0,1,-1};
char mp[1005][1005];
int vis[1005][1005]={0};
int nx,ny;
int ans;
int n,m;
int s1,s2;
void dfs(int i,int j)
{
    if(i<0||i>n||j<0||j>m)return ;
    vis[i][j]=1;
    for(int k=0;k<4;k++){
        nx=i+x[k];ny=j+y[k];
        if(nx>=0&&nx<n&&ny>=0&&ny<m&&!vis[nx][ny]&&mp[nx][ny]=='.'){ans++;vis[nx][ny]=1;dfs(nx,ny);}
    }
}
int main()
{
    while(cin>>m>>n){
            if(m==0&&n==0)break;
    getchar();
    for(int i=0;i<n;i++)
        for(int j=0;j<m;j++)
    {
        cin>>mp[i][j];
        if(mp[i][j]=='@'){s1=i;s2=j;}
    }
    memset(vis,0,sizeof(vis));
    ans=0;
    dfs(s1,s2);
    cout<<ans+1<<endl;
    }
    return 0;
}