Shichikuji and Power Grid

D. Shichikuji and Power Grid

参考：Codeforces Round #597 (Div. 2)

思路：一个很裸的最小生成树。把建立基站看成是，城市与源点（虚构的）建边。由此建立最小生成树，即可得出答案。

代码：

// Created by CAD on 2019/11/2.
#include <bits/stdc++.h>
#define ll long long
#define pii pair<int,int>
using namespace std;

const int maxn=2020*2000;
struct edge{
    ll u,v;
    ll w;
    bool operator<(edge &e)
    {
        return w<e.w;
    }
}e[maxn];
ll fa[2005],x[2005],y[2005],c[2005],k[2005];
int n;
vector<int> out1;
vector<pii> out2;
int find(int i)
{
    return i==fa[i]?i:fa[i]=find(fa[i]);
}
int cnt=0;
int cnt1=0;
ll kruskal()
{
    ll ans=0;
    for(int i=0;i<=n;++i)
        fa[i]=i;
    for(int i=1;i<=cnt;++i)
    {
        int u=find(e[i].u),v=find(e[i].v);
        if(u!=v)
        {
            ans+=e[i].w,fa[v]=u;
            if(!e[i].u) out1.push_back(e[i].v);
            else out2.push_back({e[i].u,e[i].v});
        }
    }
    return ans;
}
int main()
{
//    FOPEN;
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n;
    for(int i=1;i<=n;++i)
        cin>>x[i]>>y[i];
    for(int i=1;i<=n;++i)
        cin>>c[i];
    for(int i=1;i<=n;++i)
        cin>>k[i];
    for(int i=1;i<=n;++i)
        e[++cnt]=edge{0,i,c[i]};
    for(int i=1;i<=n;++i)
        for(int j=i+1;j<=n;++j)
            e[++cnt]=edge{i,j,1ll*(abs(x[i]-x[j])+abs(y[i]-y[j]))*(k[i]+k[j])};
    sort(e+1,e+cnt+1);
    cout<<kruskal()<<endl;
    cnt1=out1.size();
    cout<<cnt1<<endl;
    for(int i=0;i<cnt1-1;++i)
        cout<<out1[i]<<" ";
    if(cnt1)
    cout<<out1[cnt1-1]<<endl;
    cout<<out2.size()<<endl;
    for(auto i:out2)
        cout<<i.first<<" "<<i.second<<endl;
    return 0;
}