[POJ] 1979 Red and Black

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 21102		Accepted: 11267

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

Source

Japan 2004 Domestic

 

题解：广度优先搜索，从@处出发统计能遍历的black tile的个数，@也算一个。

 

代码：

#include<stdio.h>
#include<stdbool.h>
#include<string.h>
int i,j,n,m,sum,head,tail,sum,x,y,
    q[100000][2],
    dx[4]={0,0,1,-1},
    dy[4]={1,-1,0,0};

char a[100][100];

bool can[100][100];

int 
pre()
{
    memset(q,0,sizeof(q));
    memset(can,false,sizeof(can));
    sum=1;
    return 0;
}

int
init()
{
    for(i=1;i<=n;i++)
    {
        scanf("%s",&a[i]);
        for(j=0;j<m;j++)
        {
            if(a[i][j]=='.')
            can[i][j]=true;
            
            if(a[i][j]=='@')
            {
                x=i;y=j;
                can[i][j]=true;
            }
        }
    }
    
    return 0;
}

int 
bfs()
{
    int i,xi,yi,xx,yy;
    
    head=0;tail=1;
    can[x][y]=false;
    q[1][0]=x;q[1][1]=y;
    
    while(head!=tail)
    {
        head=head%100000+1;
        xi=q[head][0];
        yi=q[head][1];
        
        for(i=0;i<4;i++)
        {
            xx=xi+dx[i];
            yy=yi+dy[i];
            if(can[xx][yy])
            {
                tail=tail%100000+1;
                q[tail][0]=xx;
                q[tail][1]=yy;
                can[xx][yy]=false;
                sum++;
            }
        }
    }
    printf("%d
",sum);
    return 0;
    
    
}
             
int 
main()
{
    while(true)
    {
        pre();
        scanf("%d%d
",&m,&n);
        if(m==0&&n==0) break;
        init();
        bfs();
    }
    
   
    return 0;
}