zoukankan      html  css  js  c++  java
  • Arctic Network POJ

    The Department of National Defence (DND) wishes to connect several northern outposts by a wireless network. Two different communication technologies are to be used in establishing the network: every outpost will have a radio transceiver and some outposts will in addition have a satellite channel. 
    Any two outposts with a satellite channel can communicate via the satellite, regardless of their location. Otherwise, two outposts can communicate by radio only if the distance between them does not exceed D, which depends of the power of the transceivers. Higher power yields higher D but costs more. Due to purchasing and maintenance considerations, the transceivers at the outposts must be identical; that is, the value of D is the same for every pair of outposts. 

    Your job is to determine the minimum D required for the transceivers. There must be at least one communication path (direct or indirect) between every pair of outposts.

    Input

    The first line of input contains N, the number of test cases. The first line of each test case contains 1 <= S <= 100, the number of satellite channels, and S < P <= 500, the number of outposts. P lines follow, giving the (x,y) coordinates of each outpost in km (coordinates are integers between 0 and 10,000).

    Output

    For each case, output should consist of a single line giving the minimum D required to connect the network. Output should be specified to 2 decimal points.

    Sample Input

    1
    2 4
    0 100
    0 300
    0 600
    150 750
    

    Sample Output

    212.13

    从最小生成树中删除几条最长的边
    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<cstring>
    #include<sstream>
    #include<algorithm>
    #include<queue>
    #include<vector>
    #include<cmath>
    #include<map>
    #include<stack>
    #include<set>
    #include<memory>
    #include<bitset>
    #include<string>
    #include<functional>
    #include<iomanip>
    using namespace std;
    typedef long long LL;
    typedef unsigned long long ULL;
    #define INF 0x3f3f3f3f
    #define MAXN 509
    
    /*
    最小生成树,把所有匹配的边记录下来处理
    */
    
    struct node
    {
        double x, y;
    }a[MAXN];
    double dist(const node& a, const node& b)
    {
        return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y));
    }
    struct edge
    {
        edge(int _a, int _b, double _c) :u(_a), v(_b), cost(_c) {}
        int u, v;
        double cost;
        bool operator<(const edge& rhs)const
        {
            return cost < rhs.cost;
        }
    };
    int pre[MAXN];
    int T, k, n;
    vector<edge> E;
    vector<double> ans;
    int find(int x)
    {
        if (pre[x] == -1)
            return x;
        else
            return pre[x] = find(pre[x]);
    }void mix(int x, int y)
    {
        int fx = find(x), fy = find(y);
        if (fx != fy)
            pre[fx] = fy;
    }
    void Kruskal()
    {
        sort(E.begin(), E.end());
        for (int i = 0; i < E.size(); i++)
        {
            int f = E[i].u, t = E[i].v;
            if (find(f) != find(t))
            {
                mix(f, t);
                ans.push_back(E[i].cost);
            }
        }
    }
    int main()
    {
        ios::sync_with_stdio(false);
        cin >> T;
        while (T--)
        {
            ans.clear();
            E.clear();
            memset(pre, -1, sizeof(pre));
            cin >> k >> n;
            for (int i = 1; i <= n; i++)
            {
                cin >> a[i].x >> a[i].y;
            }
            for (int i = 1; i <= n; i++)
            {
                for (int j = i + 1; j <= n; j++)
                {
                    E.push_back(edge(i, j, dist(a[i], a[j])));
                }
            }
            Kruskal();
            sort(ans.begin(), ans.end());
            k--;
            while (k)
                k--,ans.pop_back();
            cout << setiosflags(ios::fixed); 
            cout << setprecision(2) << ans.back() << endl;    //输出0位小数,3
        }
    }
  • 相关阅读:
    tp5 宝塔open_basedir restriction in effect 错误; IIS open_basedir restriction in effect
    如何封装一个自己的win7系统并安装到电脑做成双系统
    推荐7个模板代码和其他游戏源码下载的网址
    PHP简单实现异步多文件上传并使用Postman测试提交图片
    PHP公众号开发给用户发微信消息提醒功能
    解决在页面中无法获取qrcode.js生成的base64的图片
    如何使用GUID硬盘分区格式安装新windows系统
    超详细的纯净windows系统重装示例
    Spark之join、leftOuterJoin、rightOuterJoin及fullOuterJoin
    Spark中groupByKey、reduceByKey与sortByKey
  • 原文地址:https://www.cnblogs.com/joeylee97/p/7403790.html
Copyright © 2011-2022 走看看