uva 572 Oil Deposits

原题：
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
中文：
给一个个图，@代表油田，*代表土地。如果一个坐标上，下，左，右，左上，左下，右上，右下存在油田，那么这些油田就能连成一大块。现在问你这个图里有多少个大块油田？

#include <bits/stdc++.h>using namespace std;int n,m,flag;int Move[9][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{1,-1},{-1,1},{-1,-1},{0,0}};char G[101][101];bool judge(int x,int y){    if((x>=1||x<=m)&&(y>=1||y<=n))        return true;    return false;}void dfs(int x,int y){    for(int i=0;i<9;i++)    {        int xx=x+Move[i][0];        int yy=y+Move[i][1];        if(G[xx][yy]=='@'&&judge(xx,yy))        {            G[xx][yy]='*';            flag=true;            dfs(xx,yy);        }    }}int main(){    ios::sync_with_stdio(false);    while(cin>>m>>n,m+n)    {        for(int i=1;i<=m;i++)            for(int j=1;j<=n;j++)                cin>>G[i][j];        int ans=0;        for(int i=1;i<=m;i++)        {            for(int j=1;j<=n;j++)            {                flag=false;                dfs(i,j);                if(flag)                    ans++;            }        }        cout<<ans<<endl;    }    return 0;}

解答：
这题我在杭电，北大，cf上面都做过。很基础的题目，深搜每个坐标，找8个方向的的坐标判断是否是油田，如果是就标记上’*’即可。