Oil Deposits HDU - 1241

最新推荐文章于 2021-08-22 19:31:09 发布

原创最新推荐文章于 2021-08-22 19:31:09 发布 · 450 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#Oil Deposits #HDU1241

水题同时被 2 个专栏收录

12 篇文章

订阅专栏

dfs

4 篇文章

订阅专栏

本文介绍了一种通过网格划分来探测地下油藏分布的方法。使用深度优先搜索(DFS)来确定不同油藏的数量，当两个相邻的油点被视为同一油藏的一部分时，这种方法尤为有效。网格中的每个点表示一个地块，'@'代表可能存在油藏，'*'则表示没有油藏。

题目

Oil Deposits

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30156 Accepted Submission(s): 17445

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

思路

DFS求连通块，DFS经典应用题。。。

代码

#include <bits/stdc++.h>

using namespace std;
#define MAX_N 110
int R,C;
char maze[MAX_N][MAX_N];
void solve();
void dfs(int x, int y);

int main()
{
    int n, m;
    while (cin >> m >> n && m && n) {
		R = m; C = n;
		for (int i = 0; i < m; i++) {
			while(getchar() != '\n') continue;
			for (int j = 0; j < n; j++) {
				scanf("%c", &maze[i][j]);
			}
		}
		solve();
    }
    return 0;
}
void solve() {
	int res = 0;
	for (int i = 0; i < R; i++) {
		for (int j = 0; j < C; j++) {
			if (maze[i][j] == '@') {
				dfs(i, j);
				res++;
			}
		}
	}

	cout << res << endl;
}

//int dx[] = {-1, 0, 0, 1};
//int dy[] = { 0, 1,-1, 0};
void dfs(int x, int y) {
	maze[x][y] = '*';

	for (int dx = -1; dx <= 1; dx++) {
		for (int dy = -1; dy <= 1; dy++) {
			int nx = dx + x, ny = dy + y;
			if (nx >= 0 && nx < R && ny >= 0 && ny < C && maze[nx][ny] == '@') {
				dfs(nx, ny);
			}
		}
	}
}