zoukankan      html  css  js  c++  java
  • 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:
    Bzoj1072--Scoi2007排列perm
    Bzoj1041--Haoi2008圆上的整点
    Bzoj3932--Cqoi2015任务查询系统
    HDU 1024 Max Sum Plus Plus(DP)
    HDU 1029 Ignatius and the Princess IV
    【noip模拟题】数列
    Hello World
    vue-router 进阶
    vue2.0 源码解读(二)
    vue2.0 源码解读(一)
  • 原文地址:https://www.cnblogs.com/SBSOI/p/5575045.html
Copyright © 2011-2022 走看看