算法学习：深度优先搜索（DFS）【二】

Oil Deposits

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

---

Thinking：
这题是刚才那道水池问题的升级版，只不过这次是向八个方向搜索，根据水池那题的思路，我们可以得到一种代码，还有另一种使用二维向量和循环的方法，使代码更加简洁，这里两种方法都附上。

Code1:

#include <iostream>
#define N 105
char arr[N][N];
int n,m;
using namespace std;

//递归出口判断
bool judge(char a,char b)
{
    return (arr[a][b]!='@'||a<0||b<0||a>n-1||b>m-1)?true:false;
}
void dfs(int a,int b)
{
    if(judge(a,b))return ;

    arr[a][b]='*';
    //向八个方向搜索
    dfs(a-1,b-1);
    dfs(a-1,b);
    dfs(a-1,b+1);
    dfs(a,b-1);
    dfs(a,b+1);
    dfs(a+1,b-1);
    dfs(a+1,b);
    dfs(a+1,b+1);

}
int main(void)
{
    while(cin>>n>>m &&n&&m)
    {
        int cnt=0;
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
                cin>>arr[i][j];

        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
        {
            if(arr[i][j]=='@')
            {
                cnt++;
                dfs(i,j);

            }
        }
        cout<<cnt<<endl;
    }

    return 0;
}

Code2:

#include <iostream>
#define N 105
char arr[N][N];
int n,m;
int moves[8][2]={-1,-1,0,-1,1,-1,1,0,0,1,-1,1,-1,0};
using namespace std;
bool judge(char a,char b)
{
    return (arr[a][b]!='@'||a<0||b<0||a>n-1||b>m-1)?true:false;
}
void dfs(int a,int b)
{
    if(judge(a,b))return ;
    arr[a][b]='*';

    for(int i=0;i<8;i++)
    {
        dfs(a+moves[i][0],b+moves[i][1]);
    }
}
int main(void)
{
    while(cin>>n>>m &&n&&m)
    {
        int cnt=0;
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
                cin>>arr[i][j];

        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
        {
            if(arr[i][j]=='@')
            {
                cnt++;
                dfs(i,j);

            }
        }
        cout<<cnt<<endl;
    }

    return 0;
}