HDU - 1241 Oil Deposits

最新推荐文章于 2021-08-23 22:58:21 发布

原创最新推荐文章于 2021-08-23 22:58:21 发布 · 94 阅读

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#hdu #dfs

本文介绍了一个用于探测矩形区域地下油井分布的算法。通过深度优先搜索的方式，将相邻的油井视为同一油藏，最终统计出不同油藏的数量。

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

题意：

求有多少种油井，两个相邻的油井是一种。这道题比较特殊的地方在于要向八个方向搜索。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring> 
#include<queue>
#include<stack> 
#include<string.h>
using namespace std;
int n,m;
int ss=2;
char ma[100][100];
int mm[100][100];
int nex[8][2]={{-1,1},{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1},{-1,0}};
void dfs(int x,int y)
{
	for(int k=0;k<8;k++)//开始错误是因为k==8了

	{
		int tx=x+nex[k][0];
		int ty=y+nex[k][1];
		if(tx<0||ty<0||tx>=n||ty>=m)continue;
		if(mm[tx][ty]==1){
				mm[tx][ty]=mm[x][y];
				dfs(tx,ty);
			}
	}
}

int main()
{
	while(scanf("%d%d",&n,&m)!=EOF&&m!=0&&n!=0)
	{
		ss=2;
		for(int i=0;i<n;i++)
		{
			cin>>ma[i];
			for(int j=0;j<m;j++)
			{
				if(ma[i][j]=='*')
				mm[i][j]=0;
				else if(ma[i][j]=='@')
				mm[i][j]=1;
			}
		}
		for(int i=0;i<n;i++)
			for(int j=0;j<m;j++)
			{
				if(mm[i][j]==1)
				{
					mm[i][j]=ss;
					dfs(i,j); 
					ss++;
				}
			}
			printf("%d\n",ss-2);
	}
		
	return 0;
}

这道题我把字符地图转换成了数值地图，为的是把不同的油井用不同的数标记，在main函数中遍历所有点，看看要不要dfs,dfs中对周围没标记的点进行标记，并以这一点为基础进一步搜索，直到周围没有符合条件的点时。