正题

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题目大意

有$n$个点，连边长度不超过$D$的情况下分为$S$个联通块。
求最小的$D$

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解题思路

直接$Kruskal$连边连到只剩下$S$个联通块就好了。

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$code$

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=510,M=250000;
struct node{int x,y;double w;
}a[M];
int n,m,fa[N],k,s;
double x[N],y[N];
double ans;
bool cmp(node x,node y)
{return x.w<y.w;}
int find(int x)
{return fa[x]==x?x:find(fa[x]);}
double count_dis(int a,int b)
{return sqrt((x[a]-x[b])*(x[a]-x[b])+(y[a]-y[b])*(y[a]-y[b]));}
int main()
{scanf("%d%d",&s,&n);for(int i=1;i<=n;i++)scanf("%lf%lf",&x[i],&y[i]);for(int i=1;i<n;i++)for(int j=i+1;j<=n;j++){a[++m].x=i;a[m].y=j;a[m].w=count_dis(i,j);}for(int i=1;i<=n;i++)fa[i]=i;k=n;sort(a+1,a+1+m,cmp);for(int i=1;i<=m;i++){int Fa=find(a[i].x),Fb=find(a[i].y);if(Fa!=Fb){fa[Fb]=Fa;ans=a[i].w;k--;}if(k==s) break;}printf("%.2lf",ans);
}