Red and Black——个人c++解

Problem 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)

 

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

#include<bits/stdc++.h>
using namespace std;
int a,b,sum;
char q[20][20];
void su(int x,int y)
{
    q[x][y]='#';//将走过的改为'#'
    if(x-1>=0&&q[x-1][y]=='.')//向上
    {
        sum++;
        su(x-1,y);
    }
    if(x+1<b&&q[x+1][y]=='.')//向下
    {
        sum++;
        su(x+1,y);
    }
    if(y-1>=0&&q[x][y-1]=='.')//向左
    {
        sum++;
        su(x,y-1);
    }
    if(y+1<a&&q[x][y+1]=='.')//向右
    {
        sum++;
        su(x,y+1);
    }
    //return ;利用递归后退，每走到无路可走就后退。
}
int main()
{
    int i,j,i1,j1;
    while(cin>>a>>b,a!=0&&b!=0)
    {
        sum=1;
    for(i=0;i<b;i++)
    {
        for(j=0;j<a;j++)
        {
            cin>>q[i][j];
            if(q[i][j]=='@')//标记@的位置
            {
                i1=i;
                j1=j;
            }
        }
    }
    su(i1,j1);
    cout<<sum<<endl;
    }

    return 0;
}