Uva 10369 - Arctic Network//kruskal

最新推荐文章于 2025-08-12 23:56:42 发布

mr_zys

最新推荐文章于 2025-08-12 23:56:42 发布

阅读量526

点赞数

CC 4.0 BY-SA版权

文章标签：图论

本文链接：https://blog.youkuaiyun.com/mr_zys/article/details/8891835

Uva解题报告专栏收录该内容

54 篇文章

订阅专栏

本文讨论了使用最小生成树算法解决北极哨站网络通信问题，通过两种通信技术（无线与卫星）的组合，确保至少每一对哨站之间存在通信路径。详细介绍了算法实现过程，包括求解最小生成树的Kruskal方法，以及如何根据哨站位置和通信成本计算最短连接方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Problem C: Arctic Network

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.

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). 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

Sample Output

212.13

分析：看来半天才把题意看懂，每个哨站可以有两种练习方式，用无线地或者是卫星，用卫星没有距离限制，但是卫星频道有限；用无线电有一个cost，就是要求出这个cost。

思路：一、很明显是个求最小生成树的问题，所以先求最小生成树。

二、在求最小生成树的时候，可以适当的改动一下，用kruskal的话，每次都是找当前w最大的边，这样其实不用找够p-1条边，只需要找够p-s条边。下面是代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int maxn = 505;
struct node
{
    double x;
    double y;
} G[maxn];
int u[maxn*maxn],v[maxn*maxn];
int e[maxn*maxn];
double w[maxn*maxn];
double ans[maxn*maxn];
int f[maxn];
int s,p,t,cnt;
int cmp(const int i,const int j)
{
    return w[i] < w[j];
}
int find(int x)
{
    if(x != f[x])
    {
        f[x] = find(f[x]);
    }
    return f[x];
}

double kruskal(int ok)
{
    cnt = 0;
    for(int i = 0; i < p; i++) f[i] = i;
    sort(e,e+t,cmp);
    for(int i = 0; i < t; i++)
    {
        int edge = e[i];
        int x = find(u[edge]);
        int y = find(v[edge]);
        if(x != y)
        {
            cnt++;
            f[x] = y;
            if(cnt == ok) return w[edge];
        }
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&s,&p);
        double a,b;
        for(int i = 0; i < p; i++)
        {
            scanf("%lf%lf",&a,&b);
            G[i].x = a;
            G[i].y = b;
        }
        t = 0;
        for(int i = 0; i < p; i++)
        {
            for(int j = i + 1; j < p; j++)
            {
                u[t] = i;
                v[t] = j;
                e[t] = t;
                a = G[i].x - G[j].x;
                b = G[i].y - G[j].y;
                w[t++] = sqrt(a * a + b * b);
            }
        }
        printf("%.2lf\n",kruskal(p-s));
    }
    return 0;
}