Red and Black

最新推荐文章于 2022-01-12 23:41:22 发布

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本文介绍了一个基于深度优先搜索算法的程序，该程序用于计算在一个由黑色和红色方块组成的矩形房间中，从一个指定的黑色方块出发可以到达的所有黑色方块的数量。通过递归的方式遍历所有可达的黑色方块。

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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)

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<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int m[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
int vis[25][25],ans;
int w,h;
char str[25][25];
void DFS(int x,int y){
    if(x<0||x>=h||y<0||y>=w||str[x][y]=='#'||vis[x][y])
        return;
    else
		{
	        ans++;
	        int dx,dy;
	        for(int i=0;i<4;i++)
			{
	            vis[x][y] = 1;
	            dx = x + m[i][0];
	            dy = y + m[i][1];
	            DFS(dx,dy);
	        }
    	}
}
int main() 
{
	while(scanf("%d %d",&w,&h)!=EOF&&(w||h))
	{
		getchar();
		ans=0;
		memset(vis,0,sizeof(vis));
		for(int i=0;i<h;i++)
		{
			scanf("%s",str[i]);
		}
		for(int i=0;i<h;i++)
		 	for(int j=0;j<w;j++)
		 	{
		 		if(str[i][j]=='@')
		 		DFS(i,j);
			 }
		cout<<ans<<endl;
	}
	return 0;
}