原题传送门
二分答案,然后用类似
k
r
u
s
k
a
l
kruskal
kruskal的方法验证
把
n
2
n^2
n2条边从小到大排序,加边
对于每个
m
i
d
mid
mid,如果两点之间距离
x
<
m
i
d
x<mid
x<mid,那么这两个点必须是同一部落的;如果两点之间距离
x
>
=
m
i
d
x>=mid
x>=mid,那么这两点可以不同部落
所以就看看哪些
x
<
m
i
d
x<mid
x<mid的边必须加,因为要划分成
k
k
k个部落,所以一共要加
n
−
k
n-k
n−k条边
Code:
#include <bits/stdc++.h>
#define maxn 1010
#define maxm 1000010
using namespace std;
const double eps = 1e-10;
struct node{
int x, y, dis;
}a[maxm];
int n, m, k, f[maxn], x[maxn], y[maxn];
inline int read(){
int s = 0, w = 1;
char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-')w = -1;
for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
return s * w;
}
int getfa(int k){ return f[k] == k ? k : f[k] = getfa(f[k]); }
bool cmp(node x, node y){ return x.dis < y.dis; }
bool check(double mid){
for (int i = 1; i <= n; ++i) f[i] = i;
int cnt = 0;
for (int i = 1; i <= m; ++i){
if (a[i].dis - mid > eps) break;
int x = a[i].x, y = a[i].y, s1 = getfa(x), s2 = getfa(y);
if (s1 != s2) f[s1] = s2, ++cnt;
}
return cnt <= n - k;
}
int main(){
n = read(), k = read();
for (int i = 1; i <= n; ++i) x[i] = read(), y[i] = read();
for (int i = 1; i < n; ++i)
for (int j = i + 1; j <= n; ++j)
a[++m] = (node){i, j, (x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j])};
sort(a + 1, a + 1 + m, cmp);
double l = 0, r = 14150, ans = 0;
while (fabs(r - l) > eps){
double mid = (l + r) / 2.0;
if (check(mid * mid)) ans = mid, l = mid; else r = mid;
}
printf("%.2lf\n", ans);
return 0;
}