572 - Oil Deposits

最新推荐文章于 2023-12-04 14:11:35 发布

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最新推荐文章于 2023-12-04 14:11:35 发布

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本文链接：https://blog.youkuaiyun.com/zzyafyj/article/details/39720275

GeoSurvComp公司通过地质调查技术检测地下油藏，并使用感测设备分别对每个地块进行分析，确定是否存在油气。每发现一个包含油气的地块称为一个油藏口袋。如果两个相邻的口袋属于同一油藏，则它们共同构成了一个更大的油藏。本文介绍了一种用于识别网格中不同油藏数量的方法。

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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

分析：题目实质就是求连通块的块树。

思路：每次从有石油的结点出发，用深度优先遍历思想，标志结点。每次进行遍历都能标记一个区域，区域的个数就是所求，具体可参考：数据结构基础之图部分的黑白图像部分

代码：

#include<iostream>
#include<string.h>
#include<queue>

#define  MAX 102

using namespace std;

void init_grap(int (*G)[MAX],int row,int col)
{//初始化图，用0代表*，1代表@，并给原图包裹一圈0
	int r=1,c=1;
	char flag;
	while(cin>>flag){
		if(flag=='*')G[r][c++]=0;
		else if(flag=='@') G[r][c++]=1;
		if(c==col+1) {
			++r;
			c=1;
		}
		if(r==row+1) break;
	}
}

int visited[MAX][MAX];
void DFS(int (*G)[MAX],int r,int c)
{
	if(visited[r][c] || !G[r][c]) return ;//已经访问或者(r,c)位置没有油
	visited[r][c]=1;
    DFS(G,r-1,c-1);DFS(G,r-1,c);DFS(G,r-1,c+1);
	DFS(G,r,c-1);               DFS(G,r,c+1);
	DFS(G,r+1,c-1);DFS(G,r+1,c);DFS(G,r+1,c+1);
}

void print_grap(int (*G)[MAX],int row,int col)
{//输出图，测试用
   for(int r=0;r<row+2;++r){
	   for(int c=0;c<col+2;++c){
	       cout<<G[r][c]<<" ";
	   }
	   cout<<endl;
   }
}

int main()
{
	int row,col;
	while(cin>>row>>col){
	   if(!row && !col) break;
	   int G[MAX][MAX];
	   memset(G,0,sizeof(G));
	   memset(visited,0,sizeof(visited));
	   init_grap(G,row,col);
	   int count=0;
	   for(int i=1;i<=row;++i){
		   for(int j=1;j<=col;++j){
		      if(!visited[i][j] && G[i][j]){
				  ++count;
				  DFS(G,i,j);
			  }
		   }
	   }
	   cout<<count<<endl;
	  // print_grap(G,row,col);
	}
    return 0;
}