• uva 572 oil deposits——yhx


    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.

     1 #include<cstdio>
     2 #include<cstring>
     3 int xx[8]={-1,-1,-1,0,0,1,1,1},yy[8]={-1,0,1,-1,1,-1,0,1},m,n;
     4 bool map[110][110];
     5 void dfs(int x,int y)
     6 {
     7     int i,j,k,p,q;
     8     for (i=0;i<=7;i++)
     9     {
    10       p=x+xx[i];
    11       q=y+yy[i];
    12       if (p>=1&&p<=m&&q>=1&&q<=n&&map[p][q])
    13       {
    14           map[p][q]=0;
    15           dfs(p,q);
    16       }
    17     }
    18 }
    19 int main()
    20 {
    21     int i,j,k,p,q,ans;
    22     char c,s[150];
    23     while (scanf("%d%d",&m,&n)&&m&&n)
    24     {
    25         memset(map,0,sizeof(map));
    26         gets(s);
    27         for (i=1;i<=m;i++)
    28         {
    29             gets(s+1);
    30             for (j=1;j<=n;j++)
    31               if (s[j]=='@')
    32                 map[i][j]=1;
    33         }
    34         ans=0;
    35         for (i=1;i<=m;i++)
    36           for (j=1;j<=n;j++)
    37             if (map[i][j]) 
    38             {
    39                 map[i][j]=0;
    40                 dfs(i,j);
    41                 ans++;
    42             }
    43         printf("%d
    ",ans);
    44     }
    45 } 

    每找到一个点,就将他DFS,标记掉所有周围的点。

  • 相关阅读:
    依赖注入
    Java实现一个字符串的反转
    LRU缓存介绍与实现 (Java)
    Java中HashMap遍历的两种方法(转)
    java中判断字符串是否为只包含数字
    LeakCanary 的使用遇到的弯路
    转: BAT等研发团队的技术博客
    转: android 内存检测工具 LeakCanary 说明
    转:安桌开发开源库的推荐1
    转: 技术牛人博客
  • 原文地址:https://www.cnblogs.com/SBSOI/p/5575045.html
Copyright © 2020-2023  润新知