首页 > ACM题库 > HDU-杭电 > HDU 1241 Oil Deposits-DFS-[解题报告] C++
2013
12-04

HDU 1241 Oil Deposits-DFS-[解题报告] C++

Oil Deposits

问题描述 :

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.

输入:

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.

输出:

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.

样例输入:

1 1
*
3 5
*@*@*
**@**
*@*@*
1 8
@@****@*
5 5 
****@
*@@*@
*@**@
@@@*@
@@**@
0 0 

样例输出:

0
1
2
2

 

#include<cstdio>

const int k=110;
int dir[8][2]={-1,-1,-1,0,0,-1,-1,1,1,-1,0,1,1,0,1,1};
char map[k][k];
int cnt;
int n,m;
bool isval(int x,int y)
{
	if(x<0||y<0||x>=n||y>=m)return false;
	return true;
}
void dfs(int x,int y)
{
	int i;
	for(i=0;i<8;i++)
	{
		int tx=x+dir[i][0];
		int ty=y+dir[i][1];
		if(isval(tx,ty)
			&&(map[tx][ty]=='@'))
		{
			map[tx][ty]='*';
			dfs(tx,ty);
		}
	}
}
int main()
{
	while(scanf("%d %d",&n,&m)!=EOF)
	{
		if(n==0||m==0)break;
		int i,j;
		for(i=0;i<n;i++)
		{
			getchar();
			for(j=0;j<m;j++)
			{
				scanf("%c",&map[i][j]); 
			}
		}
		cnt=0;
		for(i=0;i<n;i++)
			for(j=0;j<m;j++)
				if(map[i][j]=='@')
				{
					++cnt;
					dfs(i,j);
				}
				printf("%d\n",cnt);
	}
	return 0;
}

 


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  1. 这道题目虽然简单,但是小编做的很到位,应该会给很多人启发吧!对于面试当中不给开辟额外空间的问题不是绝对的,实际上至少是允许少数变量存在的。之前遇到相似的问题也是恍然大悟,今天看到小编这篇文章相见恨晚。