POJ 1562 Oil Deposits（DFS）

最新推荐文章于 2020-04-09 11:07:53 发布

原创最新推荐文章于 2020-04-09 11:07:53 发布 · 201 阅读

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本文介绍了一种用于探测地下油藏的算法，通过分析矩形土地网格中的油藏口袋，使用深度优先搜索确定不同油藏的数量。算法实现了对每个可能含有油藏的地块进行递归搜索，并标记已访问区域。

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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 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

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

分析：题意：找出@连通区域的个数。简单的DFS搜索，搜索方向为8个，上、右上、右、右下、下、左下、左、左上。

代码如下：

#include<stdio.h>
#include<string.h>
int n,m,next[8][2]={{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1},{-1,0},{-1,1}};//定义8个方向 
char dis[110][110];//地图大小 

void dfs(int x,int y)
{//dfs搜索 
	int xt,yt,i;
	dis[x][y]='*';//把‘@’变为‘*’ 
	for(i=0;i<8;i++)
	{//搜索8根方向 
		xt=x+next[i][0];//横坐标变化 
		yt=y+next[i][1];//纵坐标变化 
		if(xt<0||xt>=m||yt<0||yt>=n)
		continue;//超出边界范围 
		else
		{
			if(dis[xt][yt]=='@')
			dfs(xt,yt);//如果搜索到@，则继续dfs搜索 
		}
	}
	return ;
}
int main()
{
	int i,j,d;
	while(scanf("%d%d",&m,&n),m!=0)
	{
		d=0;//@连通区域个数 
		memset(dis,0,sizeof(dis));//初始化地图 
		for(i=0;i<m;i++)
		scanf("%s",dis[i]);//输入 
		for(i=0;i<m;i++)
		{
			for(j=0;j<n;j++)
			{
				if(dis[i][j]=='@')
				{//如果遇到@则进行dfs搜索 
					dfs(i,j); 
					d++;//连通区域个数加一 
				}
			}
		}
		printf("%d\n",d);//输出连通区域个数 
	}
	return 0;
}