hdu 1241 Oil Deposits （最经典的dfs）

最新推荐文章于 2022-02-25 23:47:59 发布

最新推荐文章于 2022-02-25 23:47:59 发布 · 179 阅读

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本文介绍如何使用深度优先搜索（DFS）算法检测二维网格中地下油藏的分布情况，详细阐述了输入格式、输出结果及代码实现，提供了解决类似问题的思路和实践案例。

Oil Deposits

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3087 Accepted Submission(s): 1765

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
 

说到深搜，就是所谓的dfs，我首先就是想到这一题，因为我做的第一道深搜题就是这一道，给我非常大的启发，原来深搜（dfs）就是如此！！！

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1241

我的代码：

//Oil Deposits
#include <iostream>
#include <stdio.h>
using namespace std;

char map[101][101];
int n, m, sum;

void dfs(int i, int j)
{
    //若该点不可行或越界，返回
    if(map[i][j]!='@' || i<0 || j<0 || i>=m || j>=n) return;    
    else    //否则，标记该点为不可行，再往8个方向深搜
    {
        map[i][j] = '!';
        dfs(i-1, j-1);
        dfs(i-1, j);
        dfs(i-1, j+1);
        dfs(i, j-1);
        dfs(i, j+1);
        dfs(i+1, j-1);
        dfs(i+1, j);
        dfs(i+1, j+1);
    }
}

int main()
{
    int i, j;
    while(cin>>m>>n)
    {
        if(m==0 || n==0) break;
        sum = 0;
        for(i = 0; i < m; i++)
            for(j = 0; j < n; j++)
                cin>>map[i][j];
        for(i = 0; i < m; i++)
        {
            for(j = 0; j < n; j++)
            {
                if(map[i][j] == '@')
                {
                    dfs(i, j);
                    sum++;
                }
            }
        }
        cout<<sum<<endl;
    }

    return 0;
}