HDOJ 1241 Oil Deposits (DFS)-优快云博客

本文链接：https://blog.youkuaiyun.com/shouwang_tomorrow/article/details/47815903

本文详细解读了HDOJ1241油田勘探算法问题的求解思路，通过DFS深度优先搜索算法，计算出网格中不同油田的数量。包括输入输出规范、代码实现与详细注释。

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Oil Deposits

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 18676 Accepted Submission(s): 10765

Problem Description

The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid that divides the land into numerous square plots. It then analyzes each plot separately, using sensing equipment to determine whether or not the plot contains oil. A plot containing oil is called a pocket. If two pockets are adjacent, then they are part of the same oil deposit. Oil deposits can be quite large and may contain numerous pockets. Your job is to determine how many different oil deposits are contained in a grid.

Input

The input file contains one or more grids. Each grid begins with a line containing m and n, the number of rows and columns in the grid, separated by a single space. If m = 0 it signals the end of the input; otherwise 1 <= m <= 100 and 1 <= n <= 100. Following this are m lines of n characters each (not counting the end-of-line characters). Each character corresponds to one plot, and is either `*', representing the absence of oil, or `@', representing an oil pocket.

Output

For each grid, output the number of distinct oil deposits. Two different pockets are part of the same oil deposit if they are adjacent horizontally, vertically, or diagonally. An oil deposit will not contain more than 100 pockets.

Sample Input

  
   1 1
*
3 5
*@*@*
**@**
*@*@*
1 8
@@****@*
5 5 
****@
*@@*@
*@**@
@@@*@
@@**@
0 0

Sample Output

注 - 此题为： HDOJ 1241 Oil Deposits (DFS)

思路：与 NYOJ 27 水池数目 (DFS) 思路相同，此题是向八个方向搜索记录有多少块相连的 ‘@’

已AC代码：

#include<cstdio>
#include<cstring>
char map[125][125];
int n,m;

void DFS(int x,int y)
{
	if(x<0||x>=n||y<0||y>=m)  //判断是否越界，是否可走 
		return ;
	if(map[x][y]=='*')
		return ;
	map[x][y]='*';  // 一旦搜到 @ 就标记为 *,以后都不可再走  
	
	DFS(x,y+1);  //向八个方向搜索
	DFS(x,y-1);
	DFS(x+1,y);
	DFS(x-1,y);
	DFS(x+1,y+1);
	DFS(x+1,y-1);
	DFS(x-1,y+1);
	DFS(x-1,y-1);
}


int main()
{
	int i,j;
	while(scanf("%d%d",&n,&m),n,m)
	{
		for(i=0;i<n;++i)
			scanf("%s",map[i]);
		int ans=0;
		for(i=0;i<n;++i)
		{
			for(j=0;j<=m;++j)
			{
				if(map[i][j]=='@')
				{
					DFS(i,j);   //把与（i,j）点相连的全变为 '*' 
					ans++;    //一旦搜到 '@' 数目加 1 
				}
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}