POJ 1979-Red and Black【基础DFS】

左手的ㄟ右手 2022-08-19 04:48 12阅读 0赞

Red and Black

<table style="text-align:center"> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;30000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;27684</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;15033</td> 
  </tr> 
 </tbody> 
</table>

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

45
    59
    6
    13

Source

[Japan 2004 Domestic][]

解题思路

找到@可以行动的区域，看清楚N和M谁代表行谁代表列。

#include<stdio.h>
    #include<string.h>
    #include<algorithm>
    #define INF 0x3f3f3f3f
    #define M 300
    using namespace std;
    int dx[4]={0,1,-1,0};
    int dy[4]={1,0,0,-1};
    int n,m,x,y; 
    bool vis[M][M];
    char map[M][M];
    int ans;
    void F(int xx,int yy)
    {
    	int i,j;
    	for(i=0;i<4;i++)
    	{
    		if(map[xx+dx[i]][yy+dy[i]]=='.'&&vis[xx+dx[i]][yy+dy[i]]==false&&xx+dx[i]>=0&&yy+dy[i]>=0&&xx+dx[i]<m&&yy+dy[i]<n)
    		{
    			vis[xx+dx[i]][yy+dy[i]]=true;
    			ans++;
    			//printf("%d %d\n",xx+dx[i],yy+dy[i]);
    			F(xx+dx[i],yy+dy[i]);
    		}
    	}
    }
    int main()
    {
    	while(scanf("%d%d",&n,&m)&&(n||m))
    	{	
    		memset(vis,false,sizeof(vis));
    		for(int i=0;i<m;i++)
    		{
    			getchar();
    			for(int j=0;j<n;j++)
    			{
    				scanf("%c",&map[i][j]);
    				if(map[i][j]=='@')
    				{
    					x=i,y=j;
    					vis[i][j]=true;
    				}
    			}
    		}
    		ans=0;
    		F(x,y);
    		printf("%d\n",ans+1);
    	}
    	return 0;
    }

[Japan 2004 Domestic]: http://poj.org/searchproblem?field=source&key=Japan+2004+Domestic