【HDU】1241--Oil Deposits（DFS）

最新推荐文章于 2021-01-24 13:36:02 发布

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最新推荐文章于 2021-01-24 13:36:02 发布

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文章标签： DFS hdu oj

本文链接：https://blog.youkuaiyun.com/slycoris/article/details/76468113

本文介绍了一个算法挑战，任务是通过分析由'*'和'@'组成的矩形网格来确定不同石油沉积物的数量。每个'@'代表一个可能的石油口袋，而'*'则表示不存在石油。相邻的'@'被视为同一石油沉积物的一部分。文章提供了完整的C++代码实现，使用广度优先搜索(bfs)算法遍历网格并标记已检查过的石油口袋。

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Oil Deposits

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32715 Accepted Submission(s): 19005

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

Source

Mid-Central USA 1997

#include<cstdio>
#include<cstring>
char str[105][105];
int v[105][105],n,m;
int fx[8]={-1,1,0,0,-1,-1,1,1};
int fy[8]={0,0,-1,1,-1,1,-1,1};
void bfs(int x,int y)
{
	v[x][y]=1;
	for(int i=0;i<8;i++)
	{
		int xx=x+fx[i],yy=y+fy[i];
		if(xx>=0&&xx<n&&yy>=0&&yy<m&&!v[xx][yy]&&str[xx][yy]=='@'){
			bfs(xx,yy);
		}
	}
}
int main()
{
	while(~scanf("%d%d%*c",&n,&m)&&(n!=0&&m!=0))
	{
		for(int i=0;i<n;i++)
			scanf("%s",str[i]);
		int ans=0;
		memset(v,0,sizeof(v));
		for(int i=0;i<n;i++)
			for(int j=0;j<m;j++)
			{
				if(str[i][j]=='@'&&!v[i][j])
				{
					ans++;
					bfs(i,j);
				}
			}
		printf("%d\n",ans);
	}
return 0;
}