poj 1979 Red and Black

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 22300		Accepted: 12041

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)
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
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#[email protected]#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
[email protected]
###.###
..#.#..
..#.#..
0 0

Sample Output

45
59
6
13

Source

Japan 2004 Domestic

题意：
首先这是一个图，' . '代表黑砖，‘#’代表红砖，‘@’表示主人公呆的那块黑砖。

红砖不能走。仅仅能沿着黑砖走，每次走一步。问能走到的黑砖的最大数目！

金典DFS；
注意边界，递归结束条件！

上代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
#define maxn 25
#define MM 10000000
char s[maxn][maxn]; 
int w[maxn][maxn];  //标记走过的点
int sx,sy;
int conut;         //计数
int n,m;
int xx[]={0,0,1,-1};
int yy[]={1,-1,0,0};
int dfs(int x,int y)
{
    if(s[x][y]=='#' || w[x][y])
        return 0;
        conut++;
        w[x][y]=1;
    for(int i=0;i<4;i++)
    {
        int xi=x+xx[i];
        int yi=y+yy[i];
        if(xi>=0 && xi<n && yi>=0  && yi<m)
        dfs(xi,yi);
    }
    return conut;
}
int main()
{
    while(~scanf("%d%d",&m,&n))
    {
        if(m==0 && n==0)
            break;
        memset(w,0,sizeof(w));
        conut=0;
        for(int i=0;i<n;i++)
        {
            scanf("%s",s[i]);    //逐行输入
            for(int j=0;j<m;j++)
                if(s[i][j]=='@')//标记下起点
                {
                    sx=i;     
                    sy=j;
                }
        }
      int k=dfs(sx,sy);
      printf("%d\n",k);
    }
    return 0;
}