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：
输入一个字符矩阵，‘@’代表油田，求油田的连通区域有多少个。
题解：
DFS搜索，遍历字符矩阵，遇到油田就DFS深搜，将搜过的标记为‘*’。
AC代码：
#include<iostream>
using namespace std;
char mapp[105][105];
int n,m;
int xx[]={-1,1,0,0,1,-1,-1,1}; //八个方向搜索
int yy[]={0,0,-1,1,1,-1,1,-1};
bool check(int x,int y)
{
if(x>=0&&x<n&&y>=0&&y<m&&mapp[x][y]=='@') //判断（x,y）是否越界和这个点是不是油田。
return true;
else return false;
}
void dfs(int x,int y)
{
mapp[x][y]='*'; //将油田标记，表示已经搜过了
for(int i=0;i<8;i++)
{
int fx=x+xx[i];
int fy=y+yy[i];
if(check(fx,fy))
dfs(fx,fy); //从(fx,fy)出发继续深搜，目的是将与（x，y）联通的都标记了
}
}
int main()
{
while(cin>>n>>m&&m)
{
int ans=0;
for(int i=0;i<n;i++)
cin>>mapp[i];
for(int i=0;i<n;i++) //遍历字符矩阵
{
for(int j=0;j<m;j++)
{
if(mapp[i][j]=='@') //新的区块
{
ans++;
dfs(i,j); //将与(i，j)联通的油田都标记
}
}
}
cout<<ans<<endl;
}
return 0;
}