uva 572 Oil Deposits（DFS遍历图）

最新推荐文章于 2021-06-05 16:34:03 发布

JeraKrs

最新推荐文章于 2021-06-05 16:34:03 发布

阅读量1.7k

点赞数

CC 4.0 BY-SA版权

分类专栏：数据结构-图 UVA

本文链接：https://blog.youkuaiyun.com/keshuai19940722/article/details/9671655

GRADE:D 同时被 3 个专栏收录

1021 篇文章

订阅专栏

UVA

942 篇文章

订阅专栏

算法竞赛-第六章

32 篇文章

订阅专栏

本文介绍了一种用于探测矩形土地上油田分布的算法。通过将土地划分为多个正方形地块并使用传感设备确定是否存在油田，算法能够识别独立的油田区域，并统计整个网格中不同油田的数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

题目大意：‘@’代表油田，判断给出途中有多少块油田（注意相连的是同一块，相连是八个方向）

解题思路：遍历，碰到‘@’就讲该位置和跟它相邻的‘@’全变成' * ',相当于采集掉了（防止重复计算）。

#include<stdio.h>
#include<string.h>
#define N 105
char map[N][N];
int n, m, cnt;
const int dir[8][2] = {{-1, -1}, {-1, 0}, {-1, 1}, {0, -1}, {0, 1}, {1, -1}, {1, 0}, {1, 1}};
void DFS(int a, int b){

	map[a][b] = '*';
	for (int i = 0; i < 8; i++){
		if (a + dir[i][0] < 0 || a + dir[i][0] >= n)	continue;
		if (b + dir[i][1] < 0 || b + dir[i][1] >= m)	continue;
		if (map[a + dir[i][0]][b + dir[i][1]] == '*')	continue;
		DFS(a + dir[i][0], b + dir[i][1]);
	}
}

int main(){
	while (scanf("%d%d%*c", &n, &m), n && m){
		// Init.
		memset(map, 0, sizeof(map));
		cnt = 0;

		// Read.
		for (int i = 0; i < n; i++)
			gets(map[i]);

		// Handle.
		for (int i = 0; i < n; i ++){
			for (int j = 0; j < m; j++){
				if (map[i][j] == '@'){
					DFS(i, j);
					cnt++;
				}
			}
		}

		printf("%d\n", cnt);
	}
	return 0;}