POJ 1562-Oil Deposits

最新推荐文章于 2021-03-23 21:01:56 发布

原创最新推荐文章于 2021-03-23 21:01:56 发布 · 190 阅读

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本文介绍了一种用于探测油藏分布的算法实现。该算法通过深度优先搜索（DFS）的方法，遍历网格区域来识别并计算不同油藏的数量。具体而言，从每一个含有油藏的格子出发，将其标记为已访问，并递归地检查其周围八向邻域内的油藏，直至整个油藏被完全标记。

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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 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

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：首先找到第一个' @ '的位置，把它改成‘ * ’，然后搜索与它相邻的八个位置，有‘ @ ’就改成“ * ”,直到没有“ * ”或者出界。

在找到第二个‘ @ ’，把同上述操作把与它连通的所有‘ @ ’变成‘ * ’。直到整个地图全是‘ * ’为止。

AC代码：

#include<iostream>
#include<stdio.h>
using namespace std;
char a[101][101];
int m,n;
void dfs(int i,int j)
{int i1,j1;
if(a[i][j]!='@' || i<=0 || j<=0 || i>m || j>n) return;//若出界或者不是@则返回
else{a[i][j]='*';
for(i1=i-1;i1<=i+1;i1++)
{for(j1=j-1;j1<=j+1;j1++) dfs(i1,j1);//八个方向开始继续DFS
       }
}
}
int main()
{int i,j,t;
while((scanf("%d%d",&m,&n)!=EOF)&&(m!=0)&&(n!=0))
{for(i=1;i<=m;i++)
{for(j=1;j<=n;j++) cin>>a[i][j];}
t=0;
for(i=1;i<=m;i++)
{for(j=1;j<=n;j++) if(a[i][j]=='@') //找到第一个是@的位置，开始DFS
{
                       dfs(i,j);
                       t++;
                       }
}
cout<<t<<endl;
}
return 0;
}