HDU - 1241-day4G

最新推荐文章于 2022-10-26 09:24:45 发布

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最新推荐文章于 2022-10-26 09:24:45 发布

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分类专栏：集训

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本文详细解析了一种用于探测地下油田分布的算法实现。通过构建网格模型，利用BFS（宽度优先搜索）算法来确定不同油田的具体位置和数量。算法在处理大型矩形区域时，能有效识别并计数独立的油藏沉积。代码示例展示了如何遍历网格，标记已检测到的油藏，并统计总油藏数。

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问题来源

HDU - 1241

问题描述

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.

输入

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.

输出

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.

例子

input

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

output

解题思路

bfs判断油田有多少个，八个方向判断
找到就把它消去。

AC的代码

#include <iostream>
#include<cstdio>
#include<queue>
#include<string>
using namespace std;
char aa[102][102];
int vis[102][102];
int dy[] = { -1,-1,-1,1,1,1,0,0 };
int dx[] = { -1,1,0,1,-1,0,-1,1 };
struct nn
{
	int x,y;
};
void bfs(int m,int n)
{
	queue<nn>q;
	int sum=0;
	memset(vis, 0, sizeof(vis));
	for (int f = 1; f <= m; f++)
	{
		for (int e = 1; e <= n; e++)
		{
			if (aa[f][e] == '@')
			{
				sum++;
				aa[f][e] = '*';
				nn temp;
				temp.x = f;
				temp.y = e;
				q.push(temp);
				while (!q.empty())
				{
					nn temp1 = q.front();
					q.pop();
					for (int r = 0; r < 8; r++)
					{
						int xx = temp1.x;
						int yy = temp1.y;
						if (aa[xx + dx[r]][yy + dy[r]] == '@' && !vis[xx + dx[r]][yy + dy[r]])
						{
							nn temp2;
							temp2.x = xx + dx[r];
							temp2.y = yy + dy[r];
							q.push(temp2);
							vis[xx + dx[r]][yy + dy[r]] = 1;
							aa[xx+dx[r]][yy+dy[r]] = '*';
						}
					}
				}

			}
		}


	}
	cout << sum << endl;
}

int main()
{
	int m, n;
	while (cin >> m >> n &&m)
	{
		for (int f = 1; f <= m; f++)
			for (int e = 1; e <= n; e++)
			{
				cin >> aa[f][e];
			}
		bfs(m, n);

	}
}