uva572 - Oil Deposits 油矿

最新推荐文章于 2021-05-02 17:02:24 发布

原创最新推荐文章于 2021-05-02 17:02:24 发布 · 1.4k 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#ACM #C++

=====算法相关===== 同时被 3 个专栏收录

319 篇文章

订阅专栏

+基础算法

143 篇文章

订阅专栏

+图论

35 篇文章

订阅专栏

GeoSurvComp地质调查公司负责探测地下的油田。通过在一块土地上划分多个正方形地块，使用感应设备来判断每个地块是否含有石油。一个含有石油的地块称为一个油囊。如果两个油囊相邻，则它们属于同一个油田。任务是确定网格中包含的不同油田的数量。

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

前段时间一直刷搜索，实在吃力，每题都得花上4小时+，菜鸟压力太大。。。

这题现在看来是题简单的dfs，很经典！

AC代码：

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

int cnt, m, n;
char map[200][200];
int ck[16] = {-1,-1, -1,0, -1,1, 0,-1, 0,1, 1,-1, 1,0, 1,1};

void mark (int x, int y)
{
	map[x][y] = '*'; 
	for (int i = 0; i < 17; i += 2)
		if (x + ck[i] >= 0 && x + ck[i] <= m & y + ck[i+1] >= 0 && y + ck[i+1] <= n && map[x + ck[i]][y + ck[i+1]] == '@')
			mark(x + ck[i], y + ck[i+1]);
	return;
}

int main()
{
	while (scanf("%d%d", &m, &n) && m != 0)
	{
		memset (map, 0, sizeof (map));
		for (int i = 0; i < m; i++)
			scanf("%s", map[i]);
		cnt = 0;
		for (int i = 0; i < m; i++)
			for (int j = 0; j < n; j++)
				if(map[i][j] == '@')
				{
					cnt++;
					mark(i, j);
				}
		printf("%d\n", cnt);
	}
	return 0;
}

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

Sample Output

搜索还是弱项，要加强呢！

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

Sample Output