zoukankan      html  css  js  c++  java
  • Oil Deposits (DFS)

    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

    int dir[8][2]={1,0,1,1,1,-1,-1,0,-1,1,-1,-1,0,1,0,-1};                //全方位
    char ch[100+5][100+5];
    int n,m,cnt;
    void DFS(int x,int y)
    {
        if(ch[x][y]=='@')
            ch[x][y] = '#';
        for(int i=0;i<8;i++)
        {
            int nx = dir[i][0]+x;
            int ny = dir[i][1]+y;
            if(nx>=0 && ny>=0 && nx<n && ny<m && ch[nx][ny]=='@')
                DFS(nx,ny);
        }
    
    }
    
    int main()
    {
        TXT();/*文件测试*/     TLE;/*关闭流*/
        while(cin>>n>>m&&n&&m)
        {
            cnt=0;
            for(int i=0; i<n; i++)
                for(int j=0; j<m; j++)
                    cin>>ch[i][j];
            for(int i=0; i<n; i++)
                for(int j=0; j<m; j++)
                    if(ch[i][j]=='@')
                    {
                        DFS(i,j);
                        cnt++;
                    }
            cout<<cnt<<endl;
        }
        ok;
    }
    所遇皆星河
  • 相关阅读:
    juicer 语法
    mvc 理解
    php 之 trait
    阿里P8面试官:如何设计一个扛住千万级并发的架构?
    建模
    镜像推送时出现 server gave HTTP response to HTTPS client 问题的解决方法
    git在线练习网站
    ubuntu 20.04 LTS 更换阿里云源
    Proxmox VE(Proxmox Virtual Environment)制作优盘(U盘)启动盘的教程说明方法
    KubeSphere部署Nacos集群
  • 原文地址:https://www.cnblogs.com/Shallow-dream/p/11616463.html
Copyright © 2011-2022 走看看