Red and Black HDU - 1312

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.

InputThe 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) 
OutputFor 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

思路：BFS
AC Code:

#include<iostream>
#include<cstdio>
#include<queue>
#include<utility>
#include<cstring>
using namespace std;
typedef pair<int,int> P;
char maps[][];
int vis[][];
int dx[]={,,-,},dy[]={,,,-};
int W,H,cnt;
void bfs(P p){
    queue<P> q;q.push(p);
    while(!q.empty() ){
        p=q.front() ;q.pop() ;
        int x=p.first,y=p.second;
        for(int i=;i<;i++)
        {
            int nx=x+dx[i],ny=y+dy[i];
            if(nx>=&&nx<H&&ny>=&&ny<W&&maps[nx][ny]!='#'&&vis[nx][ny]==){
                cnt++;
                vis[nx][ny]=;
                q.push(P(nx,ny));
            }
        }
    }     

}
int main(){
    while(~scanf("%d%d",&W,&H)){
        if(W==&&H==) break;
        int x,y;
        getchar();
        memset(vis,,sizeof(vis));
        for(int i=;i<H;i++){
            for(int j=;j<W;j++){
                scanf("%c",&maps[i][j]);
                if(maps[i][j]=='@') { x=i; y=j; }
            }
            getchar();
        }
        cnt=;vis[x][y]=;
        bfs(P(x,y));
        printf("%d\n",cnt);
    }
}