B - Red and Black 问题思考

布满荆棘的人生 2022-01-05 22:13 108阅读 0赞

红黑地板问题

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
    题目大意就是给定一个矩阵图形，初始位置为@，只能走. 不能走#，问最多能走多少. 
    这是一道基础的dfs，bfs水题，然而，刚接触dfs和bfs的萌新敲得就很难受，看到这题，首先反应过来的就是用bfs解决，上下左右四个方向遍历，走过的就不能再走，好吧，写了两个bfs,第一个不知道哪错了，先放上WR代码吧
    WR：把走过的点直接变为#，把要走的位置存入队列，计数，输出

#include <iostream>
    #include <cstdio>
    #include <string>
    #include <cmath>
    #include <cstring>
    #include <algorithm>
    #include <map>
    #include <set>
    #include <vector>
    #include <stack>
    #include <queue>
    #pragma warning(disable:4996)
    using namespace std;
    //char a[105][105];
    int main()
    {
        int n, m;
        while (cin >> n >> m && n != 0 && m != 0)
        {
            char a[105][105];
            memset(a,0,sizeof(a));
            queue<int>p;
            int t1, t2;
            for (int i = 0; i < m; i++)
                for (int j = 0; j < n; j++)
                {
                    cin >> a[i][j];
                    if (a[i][j] == '@')
                    {
                        t1 = i;
                        t2 = j;
                        p.push(t1);
                        p.push(t2);
    
                    }
                }
            int num = 0, x, y;
            while (p.empty() == false)
            {
                x = p.front();
                p.pop();
                y = p.front();
                p.pop();
                num++;
                //cout << num << endl;
                if (0 <= x - 1 < m && 0 <= y < n  && a[x - 1][y] == '.')//上
                {
                    p.push(x - 1);
                    p.push(y);
                    a[x - 1][y] = '#';
                }
                if (0 <= x + 1 < m && 0 <= y < n  && a[x + 1][y] == '.')//下
                {
                    p.push(x + 1);
                    p.push(y);
                    a[x + 1][y] = '#';
                }
                if (0 <= x < m && 0 <= y - 1 < n && a[x][y - 1] == '.')//左
                {
                    p.push(x);
                    p.push(y - 1);
                    a[x][y - 1] = '#';
                }
                if (0 <= x < m && 0 <= y + 1 < n && a[x][y + 1] == '.')//右
                {
                    p.push(x);
                    p.push(y + 1);
                    a[x][y + 1] = '#';
                }
            }
            cout << num << endl;
        }
        return 0;
    }

呜呜呜\`~实在不知道哪错了

下面是一种采用标记方法的做法

AC：对走过的点进行标记，用visit数组储存访问状态，结构体储存位置存入队列（结构体在bfs中非常实用），bfs基础写法，，，

#include "pch.h"
    #include <iostream>
    #include <cstdio>
    #include <string>
    #include <cmath>
    #include <cstring>
    #include <algorithm>
    #include <map>
    #include <set>
    #include <vector>
    #include <stack>
    #include <queue>
    #pragma warning(disable:4996)
    using namespace std;
    int bfs(int x, int y);
    struct point
    {
        int x, y;
    }s, s1;
    char a[25][25];
    int visit[105][105], n, m;
    queue<point>p;
    int direct[4][2] = { {-1,0},{
        1,0},{
        0,-1},{
        0,1} };
    
    bool check(int x, int y)
    {
        if (0 <= x && x < m && 0 <= y && y < n && visit[x][y]==0 && a[x][y] == '.')
            return true;
        else
            return false;
    }
    int bfs(int x, int y)
    {
        visit[x][y] = 1;
        int sum = 0;
        while (!p.empty())
        {
            s1 = p.front();
            int t1 = s1.x;
            int t2 = s1.y;
            sum++;
            p.pop();
            for (int i = 0; i < 4; i++)
            {
                s1.x = t1 + direct[i][0];
                s1.y = t2 + direct[i][1];
            
                if (check(s1.x, s1.y))
                {
                    visit[s1.x][s1.y] = 1;
                    p.push(s1);
                
                }
    
            }
        }
        return sum;
    }
    int main()
    {
        while (cin >> n >> m && n != 0 && m != 0)
        {
            memset(visit, 0, sizeof(visit));
            memset(a, 0, sizeof(a));
            int num = 1;
            for (int i = 0; i < m; i++)
                for (int j = 0; j < n; j++)
                {
                    cin >> a[i][j];
                    if (a[i][j] == '@')
                    {
                        s.x = i;
                        s.y = j;
                        p.push(s);
                        
                    }
                }
            num = bfs(s.x, s.y);
            cout << num << endl;
    
        }
        return 0;
    }

目前还没想明白哪不一样，哎，呜呜呜

转载于:https://www.cnblogs.com/858434132-qq/p/10758845.html