poj1979 Red and Black

Red and Black

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：包含@的连通区域面积。

比较笨的方法：

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<iostream>

using namespace std;

char c[25][25];

int flag[25][25];

int sum;

int n,m;

void dfs(int x,int y)

{

    if(c[x-1][y]=='.' && flag[x-1][y]==0&&x-1<=n)

    {

        sum++;

        flag[x-1][y]=1;

        dfs(x-1,y);

    }

    if(c[x][y-1]=='.' && flag[x][y-1]==0&&y-1<=m)

    {

        sum++;

        flag[x][y-1]=1;

        dfs(x,y-1);

    }

    if(c[x][y+1]=='.' && flag[x][y+1]==0&&y+1<=m)

    {

        sum++;

        flag[x][y+1]=1;

        dfs(x,y+1);

    }

    if(c[x+1][y]=='.' && flag[x+1][y]==0&&x+1<=n)

    {

        sum++;

        flag[x+1][y]=1;

        dfs(x+1,y);

    }

}

int main()

{

    int i,j,a,b;

    while(~scanf("%d%d",&m,&n))

    {

        getchar();

        sum=0;

        if(n==0&&m==0)

            break;

        memset(flag,0,sizeof(flag));

        for(i=1;i<=n;i++)

        {

            for(j=1;j<=m;j++)

            {

                scanf("%c",&c[i][j]);

                if(c[i][j]=='@')

                {

                    a=i;

                    b=j;

                    sum++;

                    flag[i][j]=1;

                }

            }

            getchar();

        }

        dfs(a,b);

        cout<<sum<<endl;

    }

    return 0;

}