POJ 1979 Red and Black 四方向棋盘搜索

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 50913		Accepted: 27001

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

题意：@是搜索起点，#不能走，.可以走，.走过一次后会变为#，问从@开始在棋盘上一共可以走几步（@起点算一步）

#include<iostream>
#include<string.h>
#include<string>
#include<algorithm>
using namespace std;
int dir[][]={{,-},{,},{,},{-,}};
string a[];
int n,m,cnt;
int check(int x,int y)
{
    if(x>=&&x<n&&y>=&&y<m&&a[x][y]!='#')
        return ;
    else
        return ;
}
void dfs(int x,int y)
{
    if(check(x,y)==)
        return ;
    else
    {
        a[x][y]='#';
        cnt++;
        for(int i=;i<;i++)
        {
            int dx,dy;
            dx=x+dir[i][];
            dy=y+dir[i][];
            dfs(dx,dy);
        }
    }
}
int main()
{
    while(cin>>m>>n&&n&&m)
    {
        for(int i=;i<n;i++)
            cin>>a[i];
        cnt=;
        for(int i=;i<n;i++)
        {
            for(int j=;j<m;j++)
            {
                if(a[i][j]=='@')
                {
                    dfs(i,j);
                    break;
                }
            }
        }
        cout<<cnt<<endl;
    }
    return ;
}