HDU-1312 Red and Black (BFS)

迈不过友情╰ 2022-05-20 09:50 115阅读 0赞

## **                                             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.

**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**

**题意描述：**  
求出与’@’四个方向相邻的所有 ’ **.** ’ 的个数。

**解题思路：**

利用广搜查找所有的图，符合条件的点，总数加一；（啊哈算法模板题）

#include<stdio.h>
    #include<string.h>
    struct note
    {
    	int x;
    	int y;
    };
    int main()
    {
    	struct note que[2500];
    	int head,tail;
    	char a[50][50];
    	int book[50][50];
    	int i,j,k,sum,mx,my,n,m,tx,ty,strx,stry;
    	int next[4][2]={0,1,
    					1,0,
    					0,-1,
    					-1,0};
    	while(scanf("%d%d",&m,&n))
    	{
    		if(n==0&&m==0)
    			break;
    		for(i=0;i<n;i++)
    			scanf("%s",a[i]);
    		for(i=0;i<n;i++)
    			for(j=0;j<m;j++)
    				if(a[i][j]=='@')
    				{
    					strx=i;
    					stry=j;
    				}
    		sum=0;
    		memset(book,0,sizeof(book));
    		head=1;
    		tail=1;
    		
    		que[tail].x=strx;
    		que[tail].y=stry;
    		tail++;
    		book[strx][stry]=1;
    		sum=1;
    			
    		while(head<tail)
    		{
    			for(k=0;k<4;k++)
    			{
    				tx=que[head].x+next[k][0];
    				ty=que[head].y+next[k][1];
    				
    				if(tx<0||tx>n-1||ty<0||ty>m-1)
    					continue;
    				if(a[tx][ty]=='.'&&book[tx][ty]==0)
    				{
    					sum++;
    					book[tx][ty]=1;
    					que[tail].x=tx;
    					que[tail].y=ty;
    					tail++;				
    				}
    			}
    			head++;
    		}
    		printf("%d\n",sum);
    	}
    	return 0;
    }