HDU 1312 Red and Black

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23404    Accepted Submission(s): 14142

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

 

 

//暴力广搜
#include <iostream>
using namespace std;
char map[25][25];
int f[25][25];
int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
int n,m,s,p;
int safe (int a,int b)
{
    if(a<=0||b<=0||a>n||b>m) return 0;
    else return 1;
}
void work(int x,int y)
{
    int c;
    for(c=0;c<4;c++)
    {
        if(map[x+dir[c][0]][y+dir[c][1]]!='#'&&safe(x+dir[c][0],y+dir[c][1]))
            f[x+dir[c][0]][y+dir[c][1]]=p+1;
    }
    return;
}
    
int main()
{
    int si,sj;
    while(cin>>m>>n)
    {
        if(n==0&&m==0) break;
        s=0;
        int i,j;
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=m;j++)
            {
                cin>>map[i][j];
                if(map[i][j]=='@') {si=i;sj=j;}
                f[i][j]=0;
            }
        }
        f[si][sj]=1;
        for(p=1;p<=500;p++)
        {
            for(i=1;i<=n;i++)
            {
                for(j=1;j<=m;j++)
                {
                    if(f[i][j]==p)
                    {
                        work(i,j);

                    }

                }
            }
        }
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=m;j++)
            {
                if(f[i][j]>0) s++;
                
            }
        }
        cout<<s<<endl;
    }
    return 0;
}