课程练习二-1011-Oil Deposits

最新推荐文章于 2019-05-20 16:37:49 发布

liu123zhen

最新推荐文章于 2019-05-20 16:37:49 发布

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分类专栏：搜索专题文章标签： acm

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本文介绍了一个用于探测地下油藏分布的算法。通过分析一个由'*'和'@'组成的二维字符数组，使用深度优先搜索（DFS）算法来确定不同油藏的数量。两个相邻的'@'代表同一油藏的一部分。

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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. <br>

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.<br>

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.<br>

Sample Input

  
  
   
   1 1<br>*<br>3 5<br>*@*@*<br>**@**<br>*@*@*<br>1 8<br>@@****@*<br>5 5 <br>****@<br>*@@*@<br>*@**@<br>@@@*@<br>@@**@<br>0 0 <br>

Sample Output

0<br>1<br>2<br>2<br>

题意：就是一个二维字符数组，里面只有* 或 @，向上、下、左、右、和对角 八个方向检索，判断是否有@，若不能够连续的
检索到@，则继续。求最多有多少组。
题目中An oil deposit will not contain more than 100 pockets.这句话，刚开始我以为是超过100个@就多计数一次呢。
思路：dfs

代码：
#include<iostream>
#include<stdlib.h>
#include<string.h>
using namespace std;
int ans = 0;
int cou = 0;
char point[102][102];
bool flag[102][102];
int m=0, n=0;
void bfs(int a,int b)
{
	/*if (cou > 100)
	{
		cou = 0;
		ans++;
	}*/
	if (point[a][b] == '@'&&!flag[a][b])
	{
		flag[a][b] = true;
		cou++;
	}
	else return;
	if (a - 1 > 0)
	{
		if (point[a - 1][b] == '@')
		{
			bfs(a - 1, b);
		}
	}
	if (a+ 1 < m)
	{
		if (point[a + 1][b] == '@')
		{
			bfs(a + 1, b);
		}
		
	}
	if (b - 1 > 0)
	{
		if (point[a][b-1] == '@')
		{
			bfs(a, b-1);
		}
	
	}
	if (b + 1 < n)
	{
		if (point[a][b+1] == '@')
		{
			bfs(a, b+1);
		}
	}
	if (a - 1 > 0&&b-1>0)
	{
		if (point[a - 1][b-1] == '@')
		{
			bfs(a - 1, b-1);
		}


	}
	if (a + 1 < m&&b+1<n)
	{
		if (point[a + 1][b+1] == '@')
		{
			bfs(a + 1, b+1);
		}
	}
	if (a - 1 > 0 && b + 1>0)
	{
		if (point[a - 1][b + 1] == '@')
		{
			bfs(a - 1, b + 1);
		}


	}
	if (a + 1 < m&&b - 1<n)
	{
		if (point[a + 1][b - 1] == '@')
		{
			bfs(a + 1, b - 1);
		}
	}
}
int main()
{
	while (1)
	{
		int m1, n1;
		cin >> m1 >> n1;
		m = m1; n = n1;
		if (m == 0 && n == 0)
			break;
		ans = 0;
		cou = 0;
	/*	for (int i = 0; i < m; i++)
			for (int j = 0; j < n; j++)
				point[i][j]='0';
		for (int i = 0; i < m; i++)
			for (int j = 0; j < n; j++)
				flag[i][j] = false;
	*/
		memset(point, 0, sizeof(point));
		memset(flag, 0, sizeof(flag));
	/*	for (int i = 0; i < m; i++)
			for (int j = 0; j < n; j++)
				cout << flag[i][j];
	*/
		for (int i = 0; i < m; i++)
			for (int j = 0; j < n; j++)
				cin >> point[i][j];
		for (int i = 0; i < m; i++)
			for (int j = 0; j < n; j++)
			{
				if (point[i][j] == '@'&&flag[i][j]==false)
				{
					ans++;
					bfs(i, j);
				}
			}
		cout << ans << endl;
	}
	system("pause");
	return 0;
}