广度搜索A - Red and Black

最新推荐文章于 2022-05-19 17:31:37 发布

zhang_nuoyan

最新推荐文章于 2022-05-19 17:31:37 发布

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博客围绕广度搜索的Red and Black问题展开。给定一个矩形房间，有红黑两种颜色的瓷砖，人站在黑瓷砖上，只能在黑瓷砖间移动。输入房间瓷砖布局等信息，要求输出人从初始瓷砖能到达的瓷砖数量，还给出了示例输入和输出。

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广度搜索A - Red and Black

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

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include <queue>
#include <stack>
using namespace std;

int n, m;
char s[22][22];
int x, y;

int ans;
int vis[22][22];

int judge(int x, int y) {
    if (x < 1 || x > m || y < 1 || y > n)
        return 0;
    return 1;
}

void bfs(int x, int y) {
    if (vis[x][y] == 1 || !judge(x, y) || s[x][y] == '#')
        return;
    vis[x][y] = 1;
    ans++;
    bfs(x + 1, y);
    bfs(x - 1, y);
    bfs(x, y + 1);
    bfs(x, y - 1);
}

int main() {
    while (cin >> n >> m) {
        ans = 0;
        memset(vis, 0 ,sizeof vis);
        if (n == 0 && m == 0)
            break;
        getchar();
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                scanf("%c", &s[i][j]);
                if (s[i][j] == '@') {
                    x = i;
                    y = j;
                }
            }
            getchar();
        }
        //for (int i = 1; i <= m; i++) {
        //    for (int j = 1; j <= n; j++) {
        //        cout << s[i][j];
        //    }
        //    cout << '\n';
        //}
        bfs(x, y);
        cout << ans << '\n';
    }
    return 0;
}