【HDU1312：Red and Black】BFS广度优先搜索C++题解

Bertha 。 2022-11-03 12:00 8阅读 0赞

## Red and Black ##

> Time Limit: 2000/1000 MS (Java/Others)  
> Memory Limit: 65536/32768 K (Java/Others)

**Problem 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.

**题目描述翻译**  
有一个长方形的房间，上面铺着正方形的瓷砖。每个瓷砖都是红色或黑色。一个男人站在一块黑色的瓷砖上。从一块瓷砖开始，他可以移动到四块相邻的瓷砖中的一块。但他不能在红砖上移动，他只能在黑色瓷砖上移动。

写一个程序来计算他通过重复上述动作所能达到的黑色瓷砖的数量。

**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)

**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 <queue>
    
    using namespace std;
    char room[23][23];
    int dir[4][2] = { 
            { -1, 0},
            { 0,  -1},
            { 1,  0},
            { 0,  1}
    };
    int Wx, Hy, num;
    #define CHECK(x, y)(x < Wx && x >= 0 && y >= 0 && y < Hy)
    struct node { 
        int x, y;
    };
    
    void BFS(int dx, int dy) { 
        num = 1;
        queue<node> q;
        node start, next;
        start.x = dx;
        start.y = dy;
        q.push(start); // 结构体的队列要压如结构体的元素
        while (!q.empty()) { 
            start = q.front();
            q.pop();
            // cout << "out" << start.x << start.y << endl;
            for (int i = 0; i < 4; i++) // 遍历四个方向
            { 
                next.x = start.x + dir[i][0];
                next.y = start.y + dir[i][1];
                if (CHECK(next.x, next.y) && room[next.x][next.y] == '.') { 
                    room[next.x][next.y] = '#';
                    num++;
                    q.push(next);
                }
            }
        }
    }
    
    int main() { 
        int x, y, dx, dy;
        while (cin >> Wx >> Hy) { 
            if (Wx == 0 && Hy == 0)
                break;
            for (y = 0; y < Hy; y++) { 
                for (x = 0; x < Wx; x++) { 
                    cin >> room[x][y];
                    if (room[x][y] == '@') { 
                        dx = x;
                        dy = y;
                    }
                }
            }
            num = 0;
            BFS(dx, dy);
            cout << num << endl;
        }
        return 0;
    }