hdu 1241 Oil Deposits （DFS入门）

最新推荐文章于 2021-01-24 13:36:02 发布

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本文介绍了一个算法挑战，旨在通过分析网格来确定不同石油沉积物的数量。输入为一系列由字符组成的网格，代表土地的不同区域，其中‘@’表示含有石油的地块，而‘*’则表示不含石油的地块。算法通过深度优先搜索来识别独立的石油沉积物。

Oil Deposits

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33901 Accepted Submission(s): 19708

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

#include <stdio.h>
char map[200][200];
int n,m;
void dfs(int x,int y)
{
	if(x<0||y<0||x>=n||y>=m)return;
	if(map[x][y]=='*')return;
	map[x][y]='*';
	dfs(x+1,y);dfs(x+1,y-1);dfs(x+1,y+1);
	dfs(x-1,y);dfs(x-1,y-1);dfs(x-1,y+1);
	dfs(x,y-1);dfs(x,y+1);
}
int main(int argc, char *argv[])
{
	while(scanf("%d %d",&n,&m),n+m)
	{
		getchar();
		int i,j;
		for(i=0;i<n;i++)
		{
			scanf("%s",map[i]);
		}
		int sum=0;
		for(i=0;i<n;i++)
		{
			for(j=0;j<m;j++)
			{
				if(map[i][j]=='@')
				{				
					dfs(i,j);
					sum++;
				}
			}
		}
		printf("%d\n",sum);	
	}
	return 0;
}

因为把@变为'“*”所以不需要标记

#include <stdio.h>	
int n,m;
char map[100][100];
void dfs(int x,int y)
{
	if(x<0||y<0||x>=n||y>=m)
	return ;
	if(map[x][y]=='*')
	return ;
	map[x][y]='*';
	dfs(x+1,y);dfs(x+1,y-1);dfs(x+1,y+1);
	dfs(x-1,y);dfs(x-1,y-1);dfs(x-1,y+1);
	dfs(x,y-1);dfs(x,y+1);	
}
int main(int argc, char *argv[])
{

	while(scanf("%d %d",&n,&m),n+m)
	{
		int i,j;
		for(i=0;i<n;i++)  
        {  
            scanf("%s",map[i]);   
		}
		int sum=0; 
		for(i=0;i<n;i++)
		{
			for(j=0;j<m;j++)
			{
				if(map[i][j]=='@')
				{
					sum++;
					dfs(i,j);
				}
			}
		}
		printf("%d\n",sum);	
	} 
	return 0;
}