Oil Deposits（DFS深搜）

Oil Deposits（DFS深搜）

题面：

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 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

are adjacent horizontally, vertically, or diagonally. An oil deposit will not contain more than 100 pockets.

题意：给一个n*m的字符型矩阵，@代表有油，*代表没油，如果两个油（上下左右或对角线相邻），代表同一个油田，每一组数据输出方阵内的油田个数

题解：这个题可以采用深搜的做法，输入数据后用两个循环遍历每一个字符，如果是油调用dfs函数搜索，同时记录加一，dfs函数的目的是找到它周围其他的油，找到后将其标记为没油*，这样就完成一个油田的判断，最后输出记录的结果即可

代码实现：

#include<iostream>
#include<cstdio>
using namespace std;

char ac[200][200];//储存字符矩阵

int m,n,k,ans = 0;// 定义m，n和re记录油田个数
int a[8] = {-1,-1,0,1,1,1,0,-1}; //标记上下左右八个方位
int b[8] = {0,1,1,1,0,-1,-1,-1};

void dfs(int x, int y)
{
	int k;
	if(x < 0 || y < 0) //不合法的话就退出循环
	  return ;
	if(ac[x][y] == '@')
	{
		ac[x][y] = '*'; //将其变为没油
		for(int k = 0; k < 8; k++)
		  dfs(x+a[k],y+b[k]); // 遍历这个油周围的油
	}
	return;
}

int main()
{
	while(scanf("%d%d",&m,&n) != EOF && (m||n)) // m为0时输入结束
	{
		int ans = 0;
		for(int i = 0; i < m; i++)
		  scanf("%s",ac[i]);
		for(int i = 0; i < m; i++)
		{
			for(int j = 0; j < n; j++) // 遍历每一个字符，判断是不是油
			{
				if(ac[i][j] == '@')
				{
					ans++;
					dfs(i,j);
				}
			}
		}
		cout << ans << endl; // 输出结果
	}
	return 0;
}