HDU-1241Oil Deposits

最新推荐文章于 2019-07-19 09:07:49 发布

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最新推荐文章于 2019-07-19 09:07:49 发布

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分类专栏： HDU ACM

本文链接：https://blog.youkuaiyun.com/qq_40829288/article/details/80648445

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本文介绍了一种用于探测地下油藏分布的算法实现。通过深度优先搜索（DFS）的方法，该算法能有效识别并计数网格中不同油藏的数量。每个油藏由多个相连的油袋构成，算法通过遍历每个单元格来确定独立油藏的总数。

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原题连接

Oil Deposits

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40410 Accepted Submission(s): 23436

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

0

1

2

2

题目要求，求八连通块的问题，求出有多少个八连块；

这个题和点击打开链接非常像，参照这个题的解法；

#include<iostream>
#include<cstring>
using namespace std;
const int maxn=100;
int n,m,cnt;
char maze[maxn][maxn+1];
void dfs(int x,int y)
{
	maze[x][y]='*';
	for(int i=-1;i<=1;i++)
		for(int j=-1;j<=1;j++)
		{
			int nx=x+i;
			int ny=y+j;
			if(nx>=0&&nx<m&&ny>=0&&ny<n&&maze[nx][ny]=='@')
				dfs(nx,ny);
		}
	return ;
}
int main()
{
	while(cin>>m>>n&&n!=0&&m!=0)
	{
		cnt=0;
		for(int i=0;i<m;i++)
			for(int j=0;j<n;j++)
				cin>>maze[i][j];
		for(int i=0;i<m;i++)
			for(int j=0;j<n;j++)
				if(maze[i][j]=='@')
				{
					cnt++;
					dfs(i,j);
				}
		cout<<cnt<<endl;
	}
	return 0;
}