原题链接

题解

对于 \((a_1,b_1),(a_2,b_2)\) 的切比雪夫距离，可以看做成 \((\frac{a_1+b_1}{2},\frac{a_1-b_1}{2}),(\frac{a_2+b_2}{2},\frac{a_2-b_2}{2})\) 的曼哈顿距离

不信你分类讨论看看

某一点到其他点的曼哈顿距离，等于所有小于该点和大于该点的x，y的坐标差之和

code

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const ll inf=1e18;

struct node
{
    ll val,id;
};

bool cmp(node a,node b)
{
    return a.val<b.val;
}

void solve()
{
    ll n;
    cin>>n;

    vector<node> heng,shu;
    vector<ll> ans(n+1,0);
    for(ll i=1;i<=n;i++)
    {
        ll x,y;
        cin>>x>>y;

        heng.push_back({x+y,i});
        shu.push_back({x-y,i});
    }

    sort(heng.begin(),heng.end(),cmp);
    sort(shu.begin(),shu.end(),cmp);

    ll sum=0;

    for(ll i=0;i<n;i++)
    {
        ll val=heng[i].val,id=heng[i].id;

        ans[id]+=i*val-sum;
        sum+=val;
    }

    sum=0;
    for(ll i=n-1;i>=0;i--)
    {
        ll val=heng[i].val,id=heng[i].id;

        ans[id]+=sum-(n-i-1)*val;
        sum+=val;
    }

    sum=0;
    for(ll i=0;i<n;i++)
    {
        ll val=shu[i].val,id=shu[i].id;

        ans[id]+=i*val-sum;
        sum+=val;
    }

    sum=0;
    for(ll i=n-1;i>=0;i--)
    {
        ll val=shu[i].val,id=shu[i].id;

        ans[id]+=sum-(n-i-1)*val;
        sum+=val;
    }

    ll res=inf;

    for(ll i=1;i<=n;i++)
    {
        res=min(res,ans[i]);
    }
    cout<<res/2<<'\n';
}
int main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int t=1;
    //cin>>t;
    while(t--) solve();
    return 0;
}