zoukankan      html  css  js  c++  java
  • Oil Deposits HDU

    Oil Deposits

     HDU - 1241 

    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. 

    InputThe 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. 
    OutputFor 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

    0
    1
    2
    2


    注意:相邻是指8个方向
     1 #include<iostream>
     2 #include<stdio.h>
     3 #include<cstring>
     4 #include<algorithm>
     5 #include<queue>
     6 
     7 using namespace std;
     8     
     9 int dx[8] = {1,-1,0,0,1,1,-1,-1};
    10 int dy[8] = {0,0,1,-1,-1,1,1,-1};
    11 char mp[102][102];
    12 int vis[102][102];
    13 int m, n;
    14 
    15 void dfs(int x, int y)
    16 {
    17     for(int i = 0; i < 8; ++i)
    18     {
    19         int xx = x + dx[i];
    20         int yy = y + dy[i];
    21         
    22         if(xx >= 0 && xx < m && yy >= 0 && yy < n && !vis[xx][yy] && mp[xx][yy] == '@')
    23         {
    24             vis[xx][yy] = 1;
    25             dfs(xx, yy);
    26         }
    27     }
    28 }
    29 
    30 
    31 int main()
    32 {
    33     std::ios::sync_with_stdio(false);
    34     while(cin >> m >> n)
    35      { 
    36          if(m == 0)
    37              break;
    38          for(int i = 0; i < m; ++i)
    39              for(int j = 0; j < n; ++j)
    40                  cin >> mp[i][j];
    41          memset(vis, 0, sizeof(vis));
    42          int ans = 0;
    43          for(int i = 0; i < m; ++i)
    44          {
    45              for(int j = 0; j < n; ++j)
    46              {
    47                  if(mp[i][j] == '@' && !vis[i][j])
    48                  {
    49                      vis[i][j] = 1;
    50                      dfs(i, j);
    51                      ans++;
    52                  }
    53                  
    54              }
    55          }
    56          cout << ans << endl;        
    57     } 
    58     
    59     
    60     
    61     return 0;
    62 }
  • 相关阅读:
    微信小程序 简单控件
    #负分小组WEEK1#一个开头&小组介绍
    CountDownLatch使用场景及分析
    Markdown基础语法
    【转】awk的使用及字符串的操作
    【开篇有益】敢问路在何方,佛曰路就在脚下
    【EntityFramework 6.1.3】个人理解与问题记录(2)
    EETOOL简介
    VMware 安装失败 “Failed to create the requested registry key Key:installer Error:1021"
    vs2010调试时为什么会出现clr.dll与mscordacwks.dll版本不匹配?
  • 原文地址:https://www.cnblogs.com/FengZeng666/p/11518395.html
Copyright © 2011-2022 走看看