HDU-1241- Oil Deposits (搜索)

最新推荐文章于 2021-03-23 21:01:56 发布

原创最新推荐文章于 2021-03-23 21:01:56 发布 · 338 阅读

0 ·

CC 4.0 BY-SA版权

搜索专栏收录该内容

3 篇文章

订阅专栏

本文介绍了一个基于深度优先搜索（DFS）算法的解决方案，用于计算给定网格中连通的石油区块数量。通过遍历每个网格单元，如果遇到代表石油的标记则进行DFS搜索并计数，最终得出不同石油区块总数。

题目链接---Oil Deposits

Oil Deposits

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37084 Accepted Submission(s): 21508

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

题意：求石油田中连通块的个数;

题解：简单的 dfs搜索

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define MAX 105
using namespace std;
char maps[MAX][MAX];
int dis[MAX][MAX];
int dx[]={1,0,0,-1,-1,1,-1,1};
int dy[]={0,1,-1,0,-1,-1,1,1};   //表示上，下，左，右，左上，左下，右上，右下八个方向 
int m,n;
void DFS(int x,int y)
{
	dis[x][y]=1;
	for(int i=0;i<8;i++)
	{
		int xx=x+dx[i];
		int yy=y+dy[i];
		if(xx>=0&&xx<m&&yy>=0&&yy<n&&!dis[xx][yy]&&maps[xx][yy]=='@')
		{
			DFS(xx,yy); 
		}
	}
} 
int main()
{
	while(cin>>m>>n,m,n)
	{
		memset(dis,0,sizeof(dis));
		int ans=0;
		for(int i=0;i<m;i++)
		{
			cin>>maps[i]; 
		}
		for(int i=0;i<m;i++)
		{
			for(int j=0;j<n;j++)
			{
				if(maps[i][j]=='@'&&!dis[i][j])
				{
					ans++;
					DFS(i,j);
				} 
			}
		}
		cout<<ans<<endl;
	}
return 0;
}