@[qiuhanlin](/user/691315) 跑一遍 Kruskal 就好了，附代码 O v O
```c
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<ctime>
#include<queue>
#include<string>
#include<bitset>
#include<cctype>
#include<cstdlib>
#include<functional>
#include<istream>
#include<sstream>
#include<streambuf>
#define ll long long
using namespace std;
const int N=200000,inf=0x3f3f3f;
struct edge
{
double head,to,w;
bool operator < (const edge &x) const
{
return w<x.w;
}
}e[N];
int x[N],y[N],f[N],s,p,cnt=0,n=0;
double ans=0;
//bool cmp(edge a,edge b)
//{
// return a.w<b.w;
//}
double dis(int i,int j)
{
return sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
}
int find1(int x)
{
return f[x]==x?x:f[x]=find1(f[x]);
}
void Kruskal()
{
for(int i=1;i<=p;i++)
f[i]=i;
sort(e+1,e+1+n);
for(int i=1;i<=n;i++)
{
// printf("%d--%d--%.2lf\n",i,cnt,e[i].w);
double x=e[i].head,y=e[i].to,z=e[i].w;
int tx=find1(x),ty=find1(y);
if(tx!=ty)
{
f[tx]=ty;
ans=max(ans,z);
cnt++;
if(cnt>=p-s)
break;
}
}
}
int main()
{
scanf("%d%d",&s,&p);
for(int i=1;i<=p;i++)
{
scanf("%d%d",&x[i],&y[i]);
for(int j=1;j<i;j++)
{
n++;
e[n]={i,j,dis(i,j)};
}
}
Kruskal();
printf("%.2f",ans);
return 0;
}
```
by wunaidedanjuan @ 2023-08-27 17:14:27
回复必关！！！
by SNXL @ 2023-08-27 17:14:53
感谢， 已关
by SNXL @ 2023-08-27 17:15:39
不用那么长
```cpp
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <bitset>
using namespace std;
const long long maxn = 501;
const long long maxm = 250001;
long long fa[maxn];
long long find(long long x) {
return x == fa[x] ? x : fa[x] = find(fa[x]);
}
struct node {
long long first, second;
double value;
}edge[maxm];
struct point {
long long x, y;
}p[maxn];
bool operator < (node a, node b) {
return a.value < b.value;
}
long long out[maxm], cnt;
bitset<maxn> vist;
long long n, m, s;
int main() {
scanf("%lld%lld", &s, &n);
for(long long i = 1 ; i<= n ; i++) {
scanf("%lld%lld", &p[i].x, &p[i].y);
fa[i] = i;
}
for(long long i = 1 ; i<= n ; i++) {
for(long long j = i+1 ; j<= n ; j++) {
edge[++m].first = i;
edge[m].second = j;
edge[m].value = (double)sqrt((1.0*p[i].x-p[j].x)*(p[i].x-p[j].x)+(1.0*p[i].y-p[j].y)*(p[i].y-p[j].y));
}
}
sort(edge+1, edge+m+1);
for(long long i = 1 ; i<= m ; i++) {
if(find(edge[i].first) == find(edge[i].second)) {
continue;
}
fa[find(edge[i].first)] = find(edge[i].second);
out[++cnt] = i;
if(cnt == n-s) {
printf("%.2lf\n", edge[i].value);
}
}
return 0;
}
by AlexSong @ 2023-08-27 17:22:58