Los wireless communication network Valley P1991 / one pass OJ 1487] [Example 2 Arctic communication network

FIG Unicom so required, considering a minimum spanning tree with the smallest possible cost. If the plus side continuously, the dispersion of the coupling point \ (PS \) a link block, the \ (S \) radio stations may be distributed at any point in each of the link block.

D and claimed herein is the radius of coverage for all points of the radius corresponds to the minimum spanning tree bottleneck. Use kruskal even side, the answer is connected \ (ps \) side of the longest bars.

Code

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<queue>
#include<algorithm>

using namespace std;

struct ne
{
    int from,to;
    double dis;
}ng[400005];//两个点，权值表示法 

struct point
{
    int x,y;
}ps[105];

int head[5005],dis[5005],f[400005];
int s,p,eg,cnt;
double ans;
bool vis[5005];

inline void merge(int n1,int n2)
{
    f[n1]=n2;
}

inline int find(int num)
{
    if(num==f[num]) return num;
    return f[num]=find(f[num]);
}

inline bool comp(ne e1,ne e2)
{
    return e1.dis<e2.dis;
}

double MST()
{
    sort(ng+1,ng+1+eg,comp);
    for(int i=1;i<=p;i++)
        f[i]=i;
    for(int i=1;i<=eg;i++)
    {
        if(find(ng[i].from)==find(ng[i].to))
            continue;
        ans=ng[i].dis;
        f[find(ng[i].from)]=find(ng[i].to);
        if(++cnt==p-s) break;
    }
    return ans;
}

int main()
{
    ios::sync_with_stdio(false);
    cin>>s>>p;
    for(int i=1;i<=p;i++)
    {
        cin>>ps[i].x>>ps[i].y;
        //cin>>ng[i].from>>ng[i].to>>ng[i].dis;
    }
    for(int i=1;i<=p;i++)
        for(int j=1;j<i;j++)
        {
            ng[++eg].from=i;
            ng[eg].to=j;
            ng[eg].dis=sqrt((ps[i].x-ps[j].x)*(ps[i].x-ps[j].x)+(ps[i].y-ps[j].y)*(ps[i].y-ps[j].y));
        }
    printf("%.2lf",MST());
    return 0;
}