poj--1979 Red and Black

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
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

Sample Output

45
59
6
13

题目大意：一个人在房间的@处，房间里铺设着红色和黑色的瓷砖，这个人只准走黑色的瓷砖，不能走红色的瓷砖
　　　　　黑色的瓷砖用‘.’表示，红色的瓷砖用'#'表示，求这个人可以走多少个瓷砖
　
题解：本题就是搜索问题可以采用DFS和BFS两种方式，本题还是采用DFS比较好，找到这个人的起始位置，从这个起始位置开始搜索
　　　直到搜索完位置。

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int W,H;
char map[21][21];
int dir[4][2]={{0,-1},{1,0},{0,1},{-1,0}};
int ans;
void DFS(int x,int y)
{
    map[x][y]='#';//说明走过了这个瓷砖
    ans++;　　//自己要增加了
    int tx,ty;
    for(int i=0;i<4;i++)
    {
        tx=x+dir[i][1];
        ty=y+dir[i][0];
        if(tx>=0 && ty>=0 && tx<H && ty<W && map[tx][ty]=='.')
        {
            DFS(tx,ty);
        }
    }
}
int main()
{
    while(scanf("%d%d",&W,&H)!=EOF)
    {
        if(W==0&&H==0)    break;
        ans=0;
        int x,y;
        for(int i=0;i<H;i++)
        {
            for(int j=0;j<W;j++)
            cin>>map[i][j];
        }
        for(int i=0;i<H;i++)
            for(int j=0;j<W;j++)
            {
                if(map[i][j]=='@')
                {
                    x=i;
                    y=j;
                }
            }
        DFS(x,y);
        printf("%d
",ans);
    }
    return 0;
}