HDU - 1241-Oil Deposits(DFS）

最新推荐文章于 2021-01-24 13:36:02 发布

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最新推荐文章于 2021-01-24 13:36:02 发布

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分类专栏： ACM比赛题目及总结

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本文介绍了一种用于探测地下油藏的算法，该算法通过分析网格中的每个地块来确定油藏位置。通过递归搜索相邻的油藏地块，算法能够准确地识别不同的油藏沉积物。示例代码展示了如何实现这一过程。

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

我的见解：利用常规方法去做，要判断是否在这个区域里会很麻烦，但我扩大范围，将油田从第一行第一列开始装进去，空出来的第0行和第0列放置其他没用的元素，判断的时候就不用判断是否在这个区域了，代码如下，非常棒！

#include<bits/stdc++.h>
#include <iostream>
using namespace std;

char dp[105][105];

void search(int a, int b)
{
	if (dp[a][b] == '@')
	{
		dp[a][b] = '*';
		search(a - 1, b - 1);
		search(a - 1, b);
		search(a - 1, b + 1);
		search(a, b - 1);
		search(a, b + 1);
		search(a + 1, b - 1);
		search(a + 1, b);
		search(a + 1, b + 1);
	}
}

int main()
{
	int m, n,i,j;
	while (cin >> m >> n)
	{
		int k = 0;
		if (m == 0)break;
		for (i = 1; i<=m;i++)
		{	
			for (j = 1; j <= n; j++)
			{
				cin >> dp[i][j];
			}
		}
		for (i = 1; i <= m; i++)
		{
			for (j = 1; j <= n; j++)
			{
				if (dp[i][j] == '@')
				{
					search(i, j);
					k++;
				}
			}
		}
		cout << k << endl;



	}
}