zoukankan      html  css  js  c++  java
  • Oil Deposits

    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
     
    #include<stdio.h>
    #include<string.h>
    int n,m,x,y,cut,sx,sy;
    char map[1001][1001];
    void dfs(int x,int y)
    {
    	if(x<0||y<0||x>=n||y>=m||map[x][y]=='*')
    	return ;	
    	map[x][y]='*';
    	 dfs(x+1,y);
    	 dfs(x-1,y);
    	 dfs(x,y+1);
    	 dfs(x,y-1);	
    	 dfs(x+1,y+1);
    	 dfs(x-1,y+1);
    	 dfs(x+1,y-1);
    	 dfs(x-1,y-1);
    }
    int main()
    {
        while(scanf("%d%d",&n,&m)!=EOF)
        {
        	if(n==0&&m==0)
        	{
        		break;
    		}
        	memset(map,0,sizeof(map));
    		cut=0;
    	for(int i=0;i<n;i++)
    	{	
    	  scanf("%s",map[i]);
    	}
    	for(int i=0;i<n;i++)
    	{
    		for( int j=0;j<m;j++)
    	     {
    	     	  
    	     		if(map[i][j]=='@')
    			{	
    				 cut++;
    				sx=i,sy=j;
    				dfs(sx,sy); 
    		     }
    		 }
    	 } 
    	 printf("%d
    ",cut);
    	}
    	return 0;
    }

    Sample Output

    0 1 2 2
    编程五分钟,调试两小时...
  • 相关阅读:
    智能佳 金刚足球机器人 竞赛机器人 智能机器人
    DIY小能手|别买电动滑板车了,咱做一台吧
    !!2016/02/22——当日买入——事后追悔,总结经验,忘记了买票的初衷!
    20160222深夜 支撑与阻力的问题,突破要不要买,回踩要不要接
    2016/2/4——昨天操作错误
    C语言 · 瓷砖铺放
    C语言 · 字符串编辑
    C语言 · 比较字符串
    C语言 · 3000米排名预测
    C语言 · 陶陶摘苹果2
  • 原文地址:https://www.cnblogs.com/kingjordan/p/12027117.html
Copyright © 2011-2022 走看看