Red and Black

最新推荐文章于 2022-01-12 23:41:22 发布

转载最新推荐文章于 2022-01-12 23:41:22 发布 · 55 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/lengxia/p/4387839.html

本文介绍了一种基于递归深度优先搜索的迷宫寻路算法，该算法可以在一个由黑色和红色方块组成的矩形房间中寻找所有可达的黑色方块。通过将问题抽象为图的遍历问题，并使用递归函数进行深度优先搜索，可以有效地解决这个问题。

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.

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>
using namespace std;
char a[100][100];

int p(char a[100][100],int x,int y)
{int l=0;
if(a[x-1][y]=='.')
{a[x-1][y]='1';p(a,x-1,y);l=1;}
if(a[x+1][y]=='.')
{a[x+1][y]='1';p(a,x+1,y);l=1;}
if(a[x][y-1]=='.')
{a[x][y-1]='1';p(a,x,y-1);l=1;}
if(a[x][y+1]=='.')
{a[x][y+1]='1';p(a,x,y+1);l=1;}
if(l==0)
return 0;
}


int main()
{int a1,a2,b1,b2,i,j;
while(cin>>a2>>a1,a2!=0)
{for(i=1;i<=a1;i++)
for(j=1;j<=a2;j++)
{cin>>a[i][j];
if(a[i][j]=='@')
{b1=i;b2=j;}
}
for(i=0;i<=a1;i++)
{a[i][a2+1]='#';
a[i][0]='#';}
for(i=0;i<=a2;i++)
{a[a1+1][i]='#';
a[0][i]='#';}
p(a,b1,b2);
int count=0;
for(i=1;i<=a1;i++)
for(j=1;j<=a2;j++)
{if(a[i][j]=='1')
count++;}

cout<<count+1<<endl;


}

return 0;}