Codeforces 1245 D. Shichikuji and Power Grid

Portal

The classic minimum spanning tree model

$ 0 $ to build a point, even an edge to the right to all other points $ x $ to $ c [x] $ sides, even the other side between any two points, while the right was $ (k_i + k_j) (\ left | x_i -x_j \ right | + \ left | y_i-y_j \ right |) $

Then $ prim $ to find a minimum spanning tree, and then think about the program output

Multi maintain a $ frm [x] $ represent the current $ dis [x] $ is the contribution from which point on the line

 

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
using namespace std;
typedef long long ll;
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(ch<'0'||ch>'9') { if(ch=='-') f=-1; ch=getchar(); }
    while(ch>='0'&&ch<='9') { x=(x<<1)+(x<<3)+(ch^48); ch=getchar(); }
    return x*f;
}
const int N=2007;
const ll INF=1e18;
int n,x[N],y[N],c[N],k[N];
inline ll cst(int i,int j) { return 1ll*(k[i]+k[j])*(abs(x[i]-x[j])+abs(y[i]-y[j])); }
ll dis[N],ans;
bool vis[N];
vector <int> bas;
vector <pair<int,int>> edg;
int frm[N];
void Prim()
{
    for(int i=1;i<=n;i++) dis[i]=c[i];
    for(int I=1;I<=n;I++)
    {
        ll mx=INF; int pos=-1;
        for(int i=1;i<=n;i++)
            if(!vis[i]&&dis[i]<mx)
                mx=dis[i],pos=i;
        vis[pos]=1; ans+=mx;
        if(!frm[pos]) bas.push_back(pos);
        else edg.push_back(make_pair(frm[pos],pos));
        for(int i=1;i<=n;i++)
            if(!vis[i]&&dis[i]>cst(pos,i))
                dis[i]=cst(pos,i),frm[i]=pos;
    }
}
int main()
{
    n=read();
    for(int i=1;i<=n;i++)
        x[i]=read(),y[i]=read();
    for(int i=1;i<=n;i++) c[i]=read();
    for(int i=1;i<=n;i++) k[i]=read();
    Prim(); printf("%lld\n",ans);
    printf("%d\n",int(bas.size()));
    for(auto x: bas) printf("%d ",x); puts("");
    printf("%d\n",int(edg.size()));
    for(auto t: edg) printf("%d %d\n",t.first,t.second);
    return 0;
}