Red and Black POJ - 1979

迷宫寻路：广度优先搜索算法解析

最新推荐文章于 2020-02-08 22:56:55 发布

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本文深入探讨了如何使用广度优先搜索(BFS)算法在一个由黑色和红色方块组成的迷宫中寻找可达路径。通过具体实例，详细介绍了BFS算法的工作原理及其在实际场景中的应用，包括如何从起点开始遍历所有可达的黑色方块，并统计可达方块的数量。

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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
struct Node
{
    int x, y;
};
int n,m;
int xx[]={0,-1,0,1};
int yy[]={1,0,-1,0};

char Map[100][100];

int cnt=0;

void bfs(int x,int y)
{
    Node now,next;
    queue<Node>Q;
    now.x=x;
    now.y=y;
    Q.push(now);
    Map[x][y]='#';
    while(!Q.empty())
    {
        now=Q.front();
        Q.pop();
        int i;
        for(i=0;i<4;i++)
        {
            next.x=now.x+xx[i];
            next.y=now.y+yy[i];
            if(next.x>=0&&next.x<m&&next.y>=0&&next.y<n&&Map[next.x][next.y]!='#'&&Map[next.x][next.y]!='#')
            {
                Q.push(next);
                Map[next.x][next.y]='#';
                cnt ++;
            }
        }
    }
}
int main()
{
    int i,j;
    while(~scanf("%d%d",&n,&m)&&(n&&m))
    {
        memset(Map,0,sizeof(Map));
        for(i=0;i<m;i++)
            scanf("%*c%s",Map[i]);

        for(i=0;i<m;i++)
        {
            for(j=0;j<n;j++)
            {
                if(Map[i][j]=='@')
                    break;
            }
            if(j<n)
                break;
        }
        cnt = 1;

        bfs(i, j);
        printf("%d\n",cnt);
    }
    return 0;
}