POJ-1562 F-Oil Deposits

最新推荐文章于 2023-12-16 19:34:19 发布

Blank_Tt

最新推荐文章于 2023-12-16 19:34:19 发布

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分类专栏： SpringTraining2 文章标签： POJ ACM C/C++ DFS

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本文介绍了一个使用深度优先搜索(DFS)算法来确定矩形区域内油藏数量的问题解决方法。通过构建网格并标记相连的油藏区域，算法能够准确计算出不同油藏的数量。示例代码展示了如何遍历每个单元格，识别油藏并进行计数。

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题目链接:

http://poj.org/problem?id=1562

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 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

题目分析：

利用DFS求连通块，利用二重循环找到当前格子的相邻8个格子

#include<iostream>
#include<memory.h>
#define N 101
using namespace std;
int m,n;
int vis[N][N];
char data[N][N];

void dfs(int x,int y,int num)
{
	if(x<0||x>=m||y<0||y>=n) return; //格子出界
	if(vis[x][y]>0||data[x][y]!='@') return;//被访问过或者不是@ 
	vis[x][y]=num;
	for(int i=-1;i<=1;i++)
	{
		for(int j=-1;j<=1;j++)
		{
			if(i!=0||j!=0)     //8个方向搜索 
			{
				dfs(x+i,y+j,num);
			}
		}
	}
}
int main(void)
{
    while(cin>>m>>n&&(m!=0||n!=0))
    {
    	int ans = 0;
    	memset(vis,0,sizeof(vis));
    	for(int i=0;i<m;i++)
		{
			for(int j=0;j<n;j++)
			{
				cin>>data[i][j];
			}
		} 
		for(int i=0;i<m;i++)
		{
			for(int j=0;j<n;j++)
			{
				if(vis[i][j]==0&&data[i][j]=='@')
				{
					dfs(i,j,++ans);
				}
			}
		}
		cout<<ans<<endl;
	    
	}
	
}