一道最小生成树模板题，这里用的Kruskal算法，把每两点就加一条边，跑一遍最小生成树即可。
#include <bits/stdc++.h>
using namespace std;
struct node{
int l , r , w;
};
node e[4000010];
int n , maxx , tot , now , ans;
int fa[2010] , a[2010] , b[2010];
int find(int x){
if(x == fa[x]) return x;
return fa[x] = find(fa[x]);
}
bool cmp(node x , node y){
return x.w < y.w;
}
int work(int i , int j){
return (a[i] - a[j]) * (a[i] - a[j]) + (b[i] - b[j]) * (b[i] - b[j]);
}
int main(){
cin >> n >> maxx;
for(int i = 1; i <= n; i++){
fa[i] = i;
cin >> a[i] >> b[i];
}
for(int i = 1; i <= n; i++)
for(int j = 1; j < i; j++){ //只用输入一半
int x = work(i , j);
if(x < maxx) continue;
tot++;
e[tot].l = i;
e[tot].r = j;
e[tot].w = x;
}
sort(e + 1 , e + tot + 1 , cmp);
for(int i = 1; i <= tot; i++){
if(now == n - 1) break;
int x = find(e[i].l) , y = find(e[i].r);
if(x == y) continue;
fa[x] = y;
now++;
ans += e[i].w;
}
if(now != n - 1) cout << -1; //边数不够
else cout << ans;
return 0;
}