1、http://acm.hdu.edu.cn/showproblem.php?pid=1241
2、DFS简单题目，好久不做题目，手生了，练练简单的。。。
题目大意：给定一个图，上边有*和@两种标记，其中@表示石油，好多@连在一起可以看成一个大的石油区，问在这个区域中有多少个石油区
3、题目：
Oil Deposits
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7952 Accepted Submission(s): 4675
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
0 1 2 2
#include<stdio.h>
#define N 105
char map[N][N];
void dfs(int a,int b)
{if(map[a][b]=='@'){map[a][b]='*';dfs(a,b+1);dfs(a+1,b+1);dfs(a+1,b);dfs(a+1,b-1);dfs(a,b-1);dfs(a-1,b-1);dfs(a-1,b);dfs(a-1,b+1);}elsereturn ;
}
int main()
{int n,m,count;while(scanf("%d%d",&m,&n)!=EOF){count=0;if(m==0)break;for(int i=0; i<=N; i++){for(int j=0; j<=N; j++)map[i][j]='*';}for(int i=1; i<=m; i++){for(int j=0; j<=n; j++)scanf("%c",&map[i][j]);}for(int i=1; i<=m; i++){for(int j=1; j<=n; j++){if(map[i][j]=='@'){dfs(i,j);count++;}}}printf("%d\n",count);}return 0;
}
/*
3 4
**@*
*@*@
@***
4 5
@**@@
*@**@
@**@*
@**@@
0 3
*/