BZOJ - 1941 最远估价

题意:求给出的所有点的最远点减最近点的最小差值

KD树的最远估价和最近估价略微不同,直接找最远垂线,反正xjb改一下就过了

#include<bits/stdc++.h>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define rrep(i,j,k) for(register int i=j;i>=k;i--)
#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])
#define print(a) printf("%lld",(ll)(a))
#define println(a) printf("%lld\n",(ll)(a))
#define printbk(a) printf("%lld ",(ll)(a))
using namespace std;
const int MAXN = 5e5+11;
const int INF = 0x7fffffff;
typedef long long ll;
ll read(){
    ll x=0,f=1;register char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int D;
struct point{
    int x[2];
    bool operator < (const point &rhs) const{
        return x[D]<rhs.x[D];
    }
};
struct KD{
    int son[MAXN][2];
    point p[MAXN],mn[MAXN],mx[MAXN];
    int root,ans,tot;
    void init(){
        ans=INF; tot=D=0;
    }
    void pu(int o){
        rep(i,0,1){
            if(son[o][i]) rep(j,0,1){
                if(mn[son[o][i]].x[j]<mn[o].x[j]) mn[o].x[j]=mn[son[o][i]].x[j];
                if(mx[son[o][i]].x[j]>mx[o].x[j]) mx[o].x[j]=mx[son[o][i]].x[j];
            }
        }
    }
    int build(int now,int l,int r){
        int mid=l+r>>1;
        tot++; son[mid][0]=son[mid][1]=0;
        D=now;nth_element(p+l,p+mid,p+r+1);//[l,r+1)
        mn[mid].x[0]=mx[mid].x[0]=p[mid].x[0];
        mn[mid].x[1]=mx[mid].x[1]=p[mid].x[1];
        if(l<mid) son[mid][0]=build(now^1,l,mid-1);
        if(r>mid) son[mid][1]=build(now^1,mid+1,r);
        pu(mid);
        return mid;
    }
    void insert(int &o,int now,point v){
        if(!o){
            o=++tot;
            p[o].x[0]=mn[o].x[0]=mx[o].x[0]=v.x[0];
            p[o].x[1]=mn[o].x[1]=mx[o].x[1]=v.x[1];
        }else{
            insert(son[o][p[o].x[now]<v.x[now]],now^1,v);
            pu(o);
        }
    }
    inline int dis(point &a,point &b){
        return abs(a.x[0]-b.x[0])+abs(a.x[1]-b.x[1]);
    }
    inline int eva(int o,point &v){
        int res=0;
        rep(i,0,1) if(v.x[i]<mn[o].x[i]||v.x[i]>mx[o].x[i]){
            if(v.x[i]<mn[o].x[i]) res+=mn[o].x[i]-v.x[i];
            else res+=v.x[i]-mx[o].x[i];
        }
        return res;
    }
    void query1(int o,point v){
        if(!o) return;
        int d1=dis(p[o],v),d2=INF,d3=INF;
        if(d1<ans&&d1!=0) ans=d1;
        if(son[o][0]) d2=eva(son[o][0],v);
        if(son[o][1]) d3=eva(son[o][1],v);
        if(d2<d3){
            if(d2<ans) query1(son[o][0],v);
            if(d3<ans) query1(son[o][1],v);
        }else{
            if(d3<ans) query1(son[o][1],v);
            if(d2<ans) query1(son[o][0],v);
        }
    }
    inline int eva2(int o,point &v){
        int res=0;
        rep(i,0,1) res+=max(abs(v.x[i]-mn[o].x[i]),abs(v.x[i]-mx[o].x[i]));
        return res;
    }
    void query2(int o,point v){
        if(!o) return;
        int d1=dis(p[o],v),d2=0,d3=0;
        if(d1>ans) ans=d1;
        if(son[o][0]) d2=eva2(son[o][0],v);
        if(son[o][1]) d3=eva2(son[o][1],v);
        if(d2>d3){
            if(d2>ans) query2(son[o][0],v);
            if(d3>ans) query2(son[o][1],v);
        }else{
            if(d3>ans) query2(son[o][1],v);
            if(d2>ans) query2(son[o][0],v);
        }
    }
    inline ll queryMin(point v){
        ans=INF;
        query1(root,v);
        return ans; 
    }
    inline ll queryMax(point v){
        ans=0;
        query2(root,v);
        return ans;
    }
}kd;
int main(){
    int n;
    while(cin>>n){
        kd.init();
        rep(i,1,n){
            kd.p[i].x[0]=read();
            kd.p[i].x[1]=read();
        }
        kd.root=kd.build(0,1,n);
        ll res=INF;
        rep(i,1,n){
            res=min(res,kd.queryMax(kd.p[i])-kd.queryMin(kd.p[i]));
        }
        println(res);
    }
    return 0;
}