题意：

在这里插入图片描述

思路：

看错题导致误入歧途，如果能早点看见翻译也不至于一天多也没想出来。
求联通的最小代价，自然的想到了能不能建边跑最小生成树。
对于两点之间比较好弄，直接 $n^2$ 暴力建边即可。对于给每个点建发电站的代价我们可以设置一个虚拟节点，将所有点向其连个边即可。
让后跑一个裸的最小生成树，注意数组要开大点以及要开 $L L$ 。

// Problem: D. Shichikuji and Power Grid
// Contest: Codeforces - Codeforces Round #597 (Div. 2)
// URL: https://codeforces.com/contest/1245/problem/D
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#include<random>
#include<cassert>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=6000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n;
int p[N],k[N];
PII a[N];
struct Edge {int a,b;LL w;bool operator < (const Edge&W) const {return w<W.w;}
}edge[N];int find(int x) {return x==p[x]? x:p[x]=find(p[x]);
}int get(int i,int j) {int dx=a[i].X-a[j].X;int dy=a[i].Y-a[j].Y;return abs(dx)+abs(dy);
}int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);cin>>n;int tot=0;for(int i=1;i<=n;i++) scanf("%d%d",&a[i].X,&a[i].Y);for(int i=1;i<=n;i++) {int x; scanf("%d",&x);edge[++tot]={0,i,x};}for(int i=1;i<=n;i++) {scanf("%d",&k[i]);for(int j=1;j<i;j++) edge[++tot]={i,j,1ll*(k[i]+k[j])*get(i,j)};}sort(edge+1,edge+1+tot);for(int i=1;i<=n;i++) p[i]=i;LL ans=0;vector<int>v1; vector<PII>v2;for(int i=1;i<=tot;i++) {int a=edge[i].a,b=edge[i].b; LL w=edge[i].w;int pa=find(a),pb=find(b);if(pa==pb) continue;if(a==0||b==0) v1.pb(max(a,b));else v2.pb({a,b});p[pa]=pb; ans+=w;}cout<<ans<<endl;printf("%d\n",v1.size()); for(auto x:v1) printf("%d ",x); puts("");printf("%d\n",v2.size()); for(auto x:v2) printf("%d %d\n",x.X,x.Y); puts("");return 0;
}
/**/