poj 1562 Oil Deposits_不见不散的结局是曲终人散

本文介绍了一种通过深度优先搜索算法解决油藏探测问题的方法。该算法能够有效地确定给定网格中不同油藏的数量，通过遍历每个单元格并标记已访问的油藏区域来实现。适用于地质调查公司进行地下油藏分布分析。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 14512		Accepted: 7911

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 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.

Sample Input

1 1
*
3 5
*@*@*
**@**
*@*@*
1 8
@@****@*
5 5 
****@
*@@*@
*@**@
@@@*@
@@**@
0 0

Sample Output

题意：求不相连的‘@’的个数，多组输入，输入位 0 0时结束；

思路：dfs遍历，遇到‘@’标记位‘*’

#include < cstdio >

#include < cstring >

int n,m,ans;

char s[105][105];

int dp[8][2]= {{1,-1},{1,0},{1,1},{0,1},{0,-1},{-1,-1},{-1,0},{-1,1}};//八个方向；

void df_s(int x,int y)

{

    int dx,dy,k;

    s[x][y]='*';//标记遍历过的坐标；

    for(k=0; k<8; k++)

    {

        dx=x+dp[k][0];

        dy=y+dp[k][1];

        if(dx >= 0&&dx < n&&dy >= 0&&dy < m&& s[dx][dy]=='@')
            df_s(dx,dy);

    }

}

int main()

{

    while(~scanf("%d%d",&n,&m))

    {

        ans=0;

        if(n+m==0)

            break;

        int i,j;

        for(i=0; i < n;i++)

            scanf("%s",s[i]);

        for(i=0; i < n;i++)

            for(j=0; j < m;j++)

            {

                if(s[i][j]=='@')

                {

                    ans++;

                    df_s(i,j);

                }

            }

        printf("%d\n",ans);

    }

    return 0;

}