zoukankan      html  css  js  c++  java
  • HDU

    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
    题解:输入图 找石油发现石油就跑dfs 把八个方向的石油全部跑到 然后用平地将其替换 计数
    #include<bits/stdc++.h>
    using namespace std;
    char maps[101][101];
    int n,m;
    void dfs(int x,int y)
    {
        if(x<0||x>=n||y<0||y>=m||maps[x][y]!='@')
        {
            return ;
        }
        else
        {
            maps[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(cin>>n>>m)
        {
            int sum=0;
            if(n==0||m==0)
            {
                break;
            }
            for(int i=0;i<n;i++)
            {
                for(int j=0;j<m;j++)
                {
                    cin>>maps[i][j];
                }
            }
            for(int i=0;i<n;i++)
            {
                for(int j=0;j<m;j++)
                {
                    if(maps[i][j]=='@')
                    {
                        dfs(i,j);
                        sum++;
                    }
                }
            }
            cout<<sum<<endl;
        }
        return 0;
    }
  • 相关阅读:
    完美解决Python与anaconda之间的冲突问题
    ansible之roles
    ansible剧本之playbook操作
    git从远程仓库拉取内容或向远程仓库上传内容
    python操作excel
    nginx基于uwsgi部署Django
    第二篇:版本控制git之分支
    笔试题92-101
    Django总结
    笔试题70-91
  • 原文地址:https://www.cnblogs.com/xuxua/p/9546007.html
Copyright © 2011-2022 走看看