Oil Deposits——在图中找油田

最新推荐文章于 2022-01-14 13:27:20 发布

原创最新推荐文章于 2022-01-14 13:27:20 发布 · 999 阅读

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#搜索 #油田 #DFS

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本文探讨了GeoSurvComp公司如何使用地质调查方法检测地下油田，并详细解释了识别油田的具体步骤，包括创建网格划分土地、单独分析每个地块以及使用感应设备判断是否含有油土。通过分析相邻地块来确定它们是否属于同一油田，最终计算出不同油田的数量。

Oil Deposits

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 $1 \le m \le 100$ and $1 \le n \le 100$ . 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

题意：给出一个m*n个图，*代表没有油田，@代表有油田，问在这块地图中一共有多少块大的油田？注意：@的八个方向有@就相当于在同一块油田中。

做了很多遍的题目了，DFS直接能过，当初学的时候还很模糊，时隔一年，随着理解的越来越深，发现这种小题可以很快就能A掉了，心情瞬间好了。废话不多说，直接上代码：

#include <stdio.h>
#include <string.h>
#include <math.h>

int fx[8] = {-1,-1,0,1,1,1,0,-1};
int fy[8] = {0,1,1,1,0,-1,-1,-1};
char mapp[106][106];

void DFS(int a,int b)
{
    mapp[a][b] = '*';
    int i;
    for(i = 0; i < 8; i++)
    {
        if(a+fx[i]>=0&&b+fy[i]>=0&&mapp[a+fx[i]][b+fy[i]] == '@')
        {
            DFS(a+fx[i],b+fy[i]);
        }
    }
}

int main()
{
    int n,m,i,j,num;
    while(~scanf("%d%d",&m,&n),m)
    {
        num = 0;
        for(i = 0; i < m; i++)
        {
            scanf("%s",mapp[i]);
        }
        for(i = 0; i < m; i++)
        {
            for(j = 0; j < n; j++)
            {
                if(mapp[i][j] == '@')
                {
                    DFS(i,j);
                    num++;
                }
            }
        }
        printf("%d\n",num);
    }
    return 0;
}