poj 1979 Red and Black（dfs）

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 Output

45
59
6
13

题意：找出@能够到达的所有点的个数
思路：dfs和递归都能做
dfs代码：

#include <iostream>
#include<cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
char s[25][25];
int a[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
int ans,n,m;
int dfs(int x,int y)
{
    for(int i=0;i<4;i++)
    {
        int tx=x+a[i][0];
        int ty=y+a[i][1];
        if(tx>=m||tx<0||ty<0||ty>=n)
            continue;
        if(s[tx][ty]=='.')
        {
            s[tx][ty]='#';
            dfs(tx,ty);
            ans++;
        }
    }
    return 0;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        if(n==0||m==0)
            break;
        int flag=1;
        int tx,ty;
        for(int i=0;i<m;i++)
        {
            scanf("%s",s[i]);
            if(flag)
            {
                for(int j=0;j<n;j++)
                {
                    if(s[i][j]=='@')
                    {
                        tx=i,ty=j;
                        flag=0;
                        break;
                    }
                }
            }
        }
        ans= 1;
        dfs(tx,ty);
        printf("%d
",ans);
    }
    return 0;
}

递归代码：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <stdio.h>
int n,m;
char data[30][30];
using namespace std;
int f(int i,int j)
{
    if(i<0||j>n-1||i>m-1||j<0)
    return 0;
    if(data[i][j]=='#')
    return 0;
    data[i][j]='#';
    return 1+f(i-1,j)+f(i+1,j)+f(i,j-1)+f(i,j+1);
}
int main()
{
    int x,y;
    while(~scanf("%d%d",&n,&m)&&(n!=0||m!=0))
    {
        getchar();
        for(int i=0;i<m;i++)
        scanf("%s",data[i]);
        for(int i=0;i<m;i++)
        for(int j=0;j<n;j++)
        if(data[i][j]=='@')
        {
            x=i;y=j;
        }
        cout<<f(x,y)<<endl;
    }
    return 0;
}