zoukankan      html  css  js  c++  java
  • Codeforces 217A. Ice Skating 搜索

    A. Ice Skating
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Bajtek is learning to skate on ice. He's a beginner, so his only mode of transportation is pushing off from a snow drift to the north, east, south or west and sliding until he lands in another snow drift. He has noticed that in this way it's impossible to get from some snow drifts to some other by any sequence of moves. He now wants to heap up some additional snow drifts, so that he can get from any snow drift to any other one. He asked you to find the minimal number of snow drifts that need to be created.

    We assume that Bajtek can only heap up snow drifts at integer coordinates.

    Input

    The first line of input contains a single integer n (1 ≤ n ≤ 100) — the number of snow drifts. Each of the following n lines contains two integers xi and yi (1 ≤ xi, yi ≤ 1000) — the coordinates of the i-th snow drift.

    Note that the north direction coinсides with the direction of Oy axis, so the east direction coinсides with the direction of the Ox axis. All snow drift's locations are distinct.

    Output

    Output the minimal number of snow drifts that need to be created in order for Bajtek to be able to reach any snow drift from any other one.

    Sample test(s)
    input
    2
    2 1
    1 2
    
    output
    1
    
    input
    2
    2 1
    4 1
    
    output
    0
    


    Bajtek只有撞到雪堆才会停下来,若两个雪堆之间横坐标或纵坐标相同则建立一条边连接两个雪堆。

    最后dfs寻找独立的图的数量

    #include <iostream>
    
    using namespace std;
    
    int n;
    int x[1111];
    int y[1111];
    bool v[1111]={0};
    
    void dfs(int i)
    {
        v[i]=true;
        for (int j=1;j<=n;j++)
        {
            if (!v[j]&&(x[i]==x[j]||y[i]==y[j]))
            {
                dfs(j);
            }
        }
    }
    
    int main()
    {
        cin>>n;
        for (int i=1;i<=n;i++)
        {
            cin>>x[i]>>y[i];
        }
        int ans=0;
        for (int i=1;i<=n;i++)
        {
            if (!v[i])
            {
                dfs(i);
                ans++;
            }
        }
        cout<<ans-1<<endl;
        return 0;
    }
    




  • 相关阅读:
    我的第一篇博客缓存显示图片
    CSLA.Net 学习 WCF服务端与客户端配置
    CSLA.Net 学习 刚接触
    [转] DevExpress 第三方控件汉化的全部代码和使用方法
    java版飞信协议实现
    [转]C#反射技术之一读取和设置类的属性
    NHibernate帮助类
    Oracle 11g 精简客户端打包 201206更新
    Mygeneration模板(NHibernate)生成,根据kdup的修改而来
    [转]TransactionScope应用
  • 原文地址:https://www.cnblogs.com/cyendra/p/3038456.html
Copyright © 2011-2022 走看看