UVa10369

Prim模板题：先将每个点之间的距离处理一下，然后随便选取一个结点作为起始结点，Prim算出所有结点到生成树的最小距离后，有S个城市是可以使用卫星通信的，由于求的是至少需要的距离，把最大的（S-1）个距离分给卫星频道，最后输出第（S-1）个大的距离即可

#include<iostream>
#include<cstdio>
#include<string.h>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<algorithm>
#include<cstring>
#include<queue>
#include<cmath>
using namespace std;
const double INF = 10000000000000.0;
const int maxn = 500005;
int n;
double d[maxn];
int visit[505];

struct point{
    double x, y;
}Point[505];

bool cmp(int a, int b)
{
    return a > b;
}

void prim(int p)
{
    memset(visit, 0, sizeof(visit));
    for (int i = 0; i < p; i++)
        d[i] = INF;
    d[0] = 0.0;

    while (1)
    {
        int v = -1;
        double m = INF;
        for (int i = 0; i < p; i++)
        {
            if (d[i] < m && visit[i] == 0)
            {
                m = d[i];
                v = i;
            }
        }

        if (v == -1)
            break;
        visit[v] = 1;

        for (int i = 0; i < p; i++)
        {
            if (visit[i]) continue;

            double len = sqrt((Point[v].x - Point[i].x) * (Point[v].x - Point[i].x)+
                              (Point[v].y - Point[i].y) * (Point[v].y - Point[i].y));

            if (len < d[i])
                d[i] = len;
        }
    }
}

int main()
{
    cin >> n;
    while (n--)
    {
        int s, p;
        cin >> s >> p;
        for (int i = 0; i < p; i++)
            scanf("%lf %lf", &Point[i].x, &Point[i].y);

        prim(p);
        sort(d, d+p, cmp);
        printf("%.2lf\n", d[s-1]);
    }

    return 0;
}