Red and Black

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本文介绍了一个基于深度优先搜索算法的迷宫寻路问题，通过在一个由黑色和红色瓷砖组成的矩形房间中寻找所有可达的黑色瓷砖，展示了如何使用递归算法解决路径探索问题。输入包括房间的尺寸和瓷砖的颜色布局，输出则是从起始位置可达的所有黑色瓷砖的数量。

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.
译文：有一个长方形的房间铺有方形瓷砖。每块瓷砖都是红色或者黑色。一个男子站在黑色的瓷砖上。从瓷砖中，他可以移动到相邻四块瓷砖中的一块。但是他不能移动到红色瓷砖上，他只能在黑色瓷砖上移动。

编写一个程序，通过重复上述动作来计算他可以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)
The end of the input is indicated by a line consisting of two zeros.
译文：输入由多个数据集组成。数据集以包含两个正整数W和H的行开始; W和H分别是x和y方向上的瓦片数量。W和H不超过20.

数据集中还有H行，每行包含W个字符。每个字符代表一个图块的颜色，如下所示。

‘.’ - 黑色瓷砖
‘＃’ - 红色瓷砖
‘@’ - 黑色瓷砖上的男人（在数据集中恰好出现一次）
输入的结尾由两个零组成的线表示。

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).
译文：对于每个数据集，您的程序应输出一行，其中包含他可以从初始图块（包括其自身）到达的图块数量。

Sample Input

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

Sample Output

C++编写：

这道题采用深度优先搜索

#include<iostream>
using namespace std;
int W,H,sum;
char room[21][21]; 
int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};
void dfs(int x,int y)
{
    sum++;
    room[x][y]='#';
    for(int i=0;i<4;i++)
    {
        int nx=x+dx[i],ny=y+dy[i];
        if(nx>=0 && nx<H && ny>=0 && ny<W && room[nx][ny]=='.')
            dfs(nx,ny);  
    }
}
int main()
{
    ios::sync_with_stdio(false);     
    while(cin>>W>>H && W!=0 && H!=0)
    {
        sum=0; 
        for(int i=0;i<H;i++)
            cin>>room[i];
        for(int i=0;i<H;i++)
        {
            for(int j=0;j<W;j++)
            {
                if(room[i][j] == '@')
                {
                    dfs(i,j);
                    break;
                }
            }
        }
        cout<<sum<<endl;
    }
}