Red and Black

Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 118 Accepted Submission(s): 92

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

 

Sample Output

45
59
6
13

 

Source

Asia 2004, Ehime (Japan), Japan Domestic

 

Recommend

Eddy

#include<bits/stdc++.h>
#define N 25
using namespace std;
int vis[N][N];
char mapn[N][N];
int dir[4][2]={
    {1,0},
    {-1,0},
    {0,1},
    {0,-1},
};
int cur=0;
int n,m;
bool ok(int x,int y)
{
    if(x<0||x>=m||y<0||y>=n||vis[x][y]||mapn[x][y]=='#')
        return false;
    return true;
}
void dfs(int x,int y)
{
    cur++;
    for(int i=0;i<4;i++)
    {
        int fx=x+dir[i][0];
        int fy=y+dir[i][1];
        if(ok(fx,fy))
        {
            vis[fx][fy]=1;
            dfs(fx,fy);
        }
    }
}
int main()
{
    //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
    while(scanf("%d%d",&n,&m)!=EOF&&n&&m)
    {
        memset(vis,0,sizeof vis);
        int x,y;
        getchar();
        for(int i=0;i<m;i++)
        {
            scanf("%s",mapn[i]);
            //cout<<mapn[i]<<endl;
            for(int j=0;j<n;j++)
            {
                if(mapn[i][j]=='@')
                {
                    x=i;
                    y=j;
                }
                //cout<<mapn[i][j];
            }
            //cout<<endl;
        }
        cur=0;
        vis[x][y]=1;
        dfs(x,y);
        printf("%d\n",cur);
    }
    return 0;
}