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本文介绍了一个简单的深度优先搜索算法，用于解决地质勘探中检测地下油田分布的问题。通过将土地划分为网格，并利用DFS算法来确定不同油田沉积物的数量。

Oil Deposits

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 68 Accepted Submission(s) : 31

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. <br>

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.<br>

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.<br>

Sample Input

1 1<br>*<br>3 5<br>*@*@*<br>**@**<br>*@*@*<br>1 8<br>@@****@*<br>5 5 <br>****@<br>*@@*@<br>*@**@<br>@@@*@<br>@@**@<br>0 0 <br>

Sample Output

0<br>1<br>2<br>2<br>

Source

Mid-Central USA 1997

题目要求：求出油田@连成片组成的区域的个数。

思路:差不多是最简单的dfs了，只需要全部搜索一遍每次搜索过的进行标记即可。

注意事项：每次记得清零。

#include<iostream>
#include<cstdio>
#include<string.h>
using namespace std;
int dir[9][3];int m,n,x,y,sum;
int J[101][101]={0};
char o[101][101];
void dfs(int a,int b)
{
    int i;
    for(i=1;i<=8;i++)
    {
        if(a<1||b<1||a>m||b>n)continue;
        if(J[a+dir[i][1]][b+dir[i][2]]==1)continue;
        if(o[a+dir[i][1]][b+dir[i][2]]=='*')continue;
        J[a+dir[i][1]][b+dir[i][2]]=1;
        dfs(a+dir[i][1],b+dir[i][2]);
    }
}
int main()
{
    dir[1][1]=-1;dir[1][2]=0;
dir[2][1]=-1;dir[1][2]=-1;
dir[3][1]=0;dir[3][2]=-1;
dir[4][1]=-1;dir[4][2]=1;
dir[5][1]=0;dir[5][2]=1;
dir[6][1]=1;dir[6][2]=1;
dir[7][1]=1;dir[7][2]=0;
dir[8][1]=1;dir[8][2]=-1;
    while(cin>>m>>n&&m)
    {
        memset(J,0,sizeof(J));
        for(int i=1;i<=m;i++)
            for(int j=1;j<=n;j++)
            cin>>o[i][j];
        sum=0;
        for(int i=1;i<=m;i++)
            for(int j=1;j<=n;j++)
            if(o[i][j]=='@'&&J[i][j]==0){J[i][j]=1;dfs(i,j);sum++;}
            cout<<sum<<endl;
    }
}