寒假训练（dfs）

最新推荐文章于 2021-05-23 13:43:43 发布

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本文介绍了一种基于深度优先搜索(DFS)的算法，用于分析地质勘探数据网格，以确定不同油藏的数量。通过将网格划分为多个正方形地块，并使用传感设备检测每个地块是否含有石油，该算法能够有效识别相邻的油藏区域，即使它们分布在水平、垂直或对角线上。

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HDU1241

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“星” 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
星
3 5
星@星@星
星星@星星
星@星@星
1 8
@@星星星星@星
5 5
星星星星@
星@@星@
星@星星@
@@@星@
@@星星@
0 0
Sample Output
0
1
2
2
问题简介：@是油，“星”是石头，输入由这些字符构成的字符矩阵，判断有多少块含油区。
问题分析：典型的DFS问题

#include <iostream>
using namespace std;
int m, n;char map[101][101];
void dfs(int x, int y)
{
	if (map[x][y] != '@' || x<=0 || y<=0 || x>m || y>n)return;
	else {
		map[x][y] = '!';//同搜索到的@变成！然后往八个方向搜索。
		dfs(x, y - 1);
		dfs(x, y +1);
		dfs(x+1, y - 1);
		dfs(x+1, y);
		dfs(x+1, y+1 );
		dfs(x-1, y-1 );
		dfs(x-1, y +1);
		dfs(x-1, y );
	}
}
int main()
{
	while (cin >> m >> n)
	{
		if (m == 0)break;
		for (int i = 1; i <= m; i++)
		{
			for (int j = 1; j <= n; j++)
			{
				cin >> map[i][j];
			}

		}
		int ans = 0;
		for (int i = 1; i <= m; i++)
		{
			for (int j = 1; j <= n; j++)
			{
				if (map[i][j] == '@') { dfs(i, j); ans++; }
			}

		}
		cout << ans << endl;
	}
}