Red and Black

问题描述

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. 

输入

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. 

输出

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).

样例输入

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

样例输出

45
59
6
13

经典搜索算法BFS。用一个队列保存当前可以到达的位置，每存入一个位置就将这个位置标记为已经遍历过，然后每次从队列pop一个位置，并对该位置扩展可到达位置存入到队列中，直到队列为空。

#include <iostream>
#include <queue>
using namespace std;
char map[25][25];
int dx[8]={-1,0,1,0,1,-1,-1,1},
    dy[8]={0,-1,0,1,1,-1,1,-1};
void dfs(int x,int y);
int toatl=0,m,n;
typedef struct Pos{
    int x;
    int y;
}P;
queue<P> que;
P start;
int main() {
    while(cin>>n>>m) {
        if(n==0&&m==0)
            break;
        toatl=0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                cin >> map[i][j];
                if (map[i][j] == '@') {
                    start.x = i;
                    start.y = j;
                }
            }
        }
        que.push(start);

        while(que.size()){
            P t=que.front();
            que.pop();
            toatl++;

            for(int i=0;i<4;i++){
                P tmp;
                tmp.x=t.x+dx[i];
                tmp.y=t.y+dy[i];
                if(tmp.x>=0&&tmp.x<m&&tmp.y>=0&&tmp.y<n&&map[tmp.x][tmp.y]=='.'){
                    que.push(tmp);
                    /* 标记已经遍历过 */
                    map[tmp.x][tmp.y]='*';
                }
            }
        }
        cout<<toatl<<endl;
    }

    return 0;
}