HDU-1241-Oil Deposits

最新推荐文章于 2019-05-20 16:37:49 发布

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最新推荐文章于 2019-05-20 16:37:49 发布

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分类专栏：查找DFS BFS

本文链接：https://blog.youkuaiyun.com/lemon25/article/details/76472166

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本文介绍了一个用于检测地下油藏的算法。通过分析矩形区域土地上的网格数据，使用广度优先搜索(BFS)和深度优先搜索(DFS)来确定不同油藏的数量。详细解释了如何处理输入数据并提供两种搜索算法的实现。

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Oil Deposits

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32726 Accepted Submission(s): 19013

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

/******************************
 bfs(广度搜索，用于找出最优解)
*******************************/
#include<stdio.h>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
bool flag[1000][1000];
char a[1000][1000];
int fx[]={-1,-1,0,0,1,1,1,-1},fy[]={0,1,1,-1,1,-1,0,-1};  //坐标表示移动方向
int m,n;
struct node{
	int x,y;
};
void bfs(int x,int y)
{	
	int i,j;
	flag[x][y]=1;
	queue <node> qu;
	node s;
	s.x=x,s.y=y;
	qu.push(s);
	while(!qu.empty())
	{
		s=qu.front();
		qu.pop();
	
		for(i=0;i<8;i++)
		{
			int xx=s.x+fx[i],yy=s.y+fy[i];
			if(xx>=0&&yy>=0&&xx<m&&yy<n&&!flag[xx][yy]&&a[xx][yy]=='@')
			{
				node w;
				flag[xx][yy]=1;
				w.x=xx,w.y=yy;
				qu.push(w);
			}
		}
 	}
}
int main()
{
	int i,j,ans;
	
	while(~scanf("%d%d",&m,&n))
	{
		if(m==0)
		break;
		memset(flag,0,sizeof(flag));
		ans=0;
		for(i=0;i<m;i++)
			scanf("%s",a[i]);
				for(i=0;i<m;i++)
					for(j=0;j<n;j++)
				if(!flag[i][j]&&a[i][j]=='@')
			ans++,bfs(i,j);
		printf("%d\n",ans);
	}	
	return 0;
}

/*****************************
dfs(深度搜索，遍历所有解)
*****************************/
#include<stdio.h>
#include<algorithm>
#include<cstring>
using namespace std;
char a[1000][1000];
bool flag[1000][1000];
int n,m;
int fx[]={-1,-1,-1,0,0,1,1,1},fy[]={1,0,-1,1,-1,1,0,-1};
void dfs(int x,int y)
{
	int i,j;
	flag[x][y]=1;
	for(i=0;i<8;i++)
	{
		int xx=x+fx[i],yy=y+fy[i];
		if(xx>=0&&yy>=0&&xx<m&&yy<n&&!flag[xx][yy]&&a[xx][yy]=='@')
		dfs(xx,yy);
	}
}

int main()
{
	while(~scanf("%d%d",&m,&n),m)
	{
		int i,j,ans=0;
		memset(flag,0,sizeof(flag));
		for(i=0;i<m;i++)
		scanf("%s",a[i]);
		for(i=0;i<m;i++)
			for(j=0;j<n;j++)
				if(!flag[i][j]&&a[i][j]=='@')
			ans++,dfs(i,j);
		printf("%d\n",ans);	
	}
	return 0;
}