L - Oil Deposits //基础深度优先搜索，把每一个点都进去搜索一遍

L - Oil Deposits

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

思路：要找到所有种类的油田，那么判断是否为同一个油田的条件为：上下左右 交叉都算是相邻的两个油田为
同一种油，否则就是不同种类的一种油
就我个人看来我觉得应该用dfs，深度搜索，找到所有的情况

把每一个点都放入dfs中去搜索，如果不是油田的话， 就不用搜索了，如果是油田就去搜索，并且把搜索过的油田标记一下，以后就不用在搜索了，搜索油田的方向有8个方向，上下左右交叉，总共八个

代码中的flag是来判断是否是同一个油田的标记，如果不是油田就会返回为0，如果是油田的话，搜索完与当前油田相邻的油田之后，就会返回1，这样油田的种类就可以加1，这个题目比较简单，就是基础的深度优先搜索。

AC代码

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;

char mp[200][200];
int cnt;
int flag;
int vis[200][200];
int n,m;
int dir[8][2]={0,1,1,0,0,-1,-1,0,1,1,1,-1,-1,1,-1,-1};
int dfs(int x,int y)
{
	if(mp[x][y]=='*')
	  return flag;
	mp[x][y]='*';
	flag=1;
	for(int i=0;i<8;i++)
	{
		int a=x+dir[i][0];
		int b=y+dir[i][1];
		if(a>=0&&a<n&&b>=0&&b<m)
		{
		   dfs(a,b);
		}
	}
	return flag;
} 
 int main()
 {
 	while(scanf("%d%d",&n,&m)!=EOF)
 	{
 		cnt=0;
 		if(n==0 ||m==0)
 		{
 			break;
		 }
		for(int i=0;i<n;i++)
		scanf("%s",mp[i]);
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<m;j++)
			{
				flag=0;
				if(dfs(i,j))
				{
					cnt++;
				}
			}
		}
		printf("%d\n",cnt);
	 }
	 return 0;
 }