老套路了。

用二分求答案，judge时把每个点转换成矩形，最后看n个矩形是否有交点

#include<bits/stdc++.h>
using namespace std;
#define N 3005
#define ll long long

struct Point{ll x,y;}c;
struct Rec{ll x1,x2,y1,y2;}rec;
ll n,x[N],y[N],h[N];

Rec merge(Rec a,Rec b){
    Rec res;
    res.x1=max(a.x1,b.x1);
    res.y1=max(a.y1,b.y1);
    res.x2=min(a.x2,b.x2);
    res.y2=min(a.y2,b.y2);
    return res;
}

int judge(ll H){//金字塔高度为H
    Rec now;
    ll d=H-h[1];
    now=(Rec){x[1]-d,x[1]+d,y[1]-d,y[1]+d};
    for(int i=2;i<=n;i++){
        ll d=H-h[i];
        Rec t=(Rec){x[i]-d,x[i]+d, y[i]-d,y[i]+d};
        now=merge(now,t);
        if(now.x1>now.x2 || now.y1>now.y2)return 0;
    }
    c=(Point){now.x1,now.y1};
    return 1;
}

int main(){
    ll L=0,R=1e18,mid,ans;
    cin>>n;
    for(int i=1;i<=n;i++)
        scanf("%lld%lld%lld",&x[i],&y[i],&h[i]),L=max(L,h[i]);

    while(L<=R){
        mid=L+R>>1;
        if(judge(mid))
            ans=mid,R=mid-1;
        else L=mid+1;;
    }
    
    cout<<c.x<<" "<<c.y<<" "<<ans<<'\n';
}
/*
5
3 7 6
1 2 9
3 3 4 
6 4 8
12 4 1
*/