题目：http://www.lydsy.com/JudgeOnline/problem.php?id=1941
题解：CDQ神马的都是浮云！我会暴力K-Dtree我自豪！
差点进了第一页。。。好久没写K-Dtree结果没附初值调了好久T_T
代码：
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<iostream> 7 #include<vector> 8 #include<map> 9 #include<set> 10 #include<queue> 11 #include<string> 12 #define inf 1000000000 13 #define maxn 200000+5 14 #define maxm 100000+5 15 #define eps 1e-10 16 #define ll double 17 #define pa pair<ll,int> 18 #define for0(i,n) for(int i=0;i<=(n);i++) 19 #define for1(i,n) for(int i=1;i<=(n);i++) 20 #define for2(i,x,y) for(int i=(x);i<=(y);i++) 21 #define for3(i,x,y) for(int i=(x);i>=(y);i--) 22 #define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go) 23 #define mod 1000000007 24 using namespace std; 25 inline int read() 26 { 27 int x=0,f=1;char ch=getchar(); 28 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 29 while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();} 30 return x*f; 31 } 32 int n,m,cur,ans1,ans2; 33 struct rec 34 { 35 int mi[2],mx[2],d[2],l,r; 36 int& operator [](int i){return d[i];} 37 }p[maxn],t[maxn],now; 38 bool operator <(rec a,rec b){return a[cur]<b[cur];} 39 inline void pushup(int k) 40 { 41 int l=t[k].l,r=t[k].r; 42 for0(i,1) 43 { 44 t[k].mi[i]=min(t[k][i],min(t[l].mi[i],t[r].mi[i])); 45 t[k].mx[i]=max(t[k][i],max(t[l].mx[i],t[r].mx[i])); 46 } 47 } 48 inline int build(int l,int r,int dir) 49 { 50 int mid=(l+r)>>1; 51 cur=dir; 52 nth_element(p+l,p+mid,p+r+1); 53 t[mid]=p[mid]; 54 for0(i,1)t[mid].mi[i]=t[mid].mx[i]=t[mid][i]; 55 t[mid].l=l>mid-1?0:build(l,mid-1,dir^1); 56 t[mid].r=mid+1>r?0:build(mid+1,r,dir^1); 57 pushup(mid); 58 return mid; 59 } 60 inline int dist(rec a,rec b){return abs(a[0]-b[0])+abs(a[1]-b[1]);} 61 inline int calc1(int k) 62 { 63 if(!k)return inf; 64 int ret=0; 65 for0(i,1) 66 { 67 if(now[i]<t[k].mi[i])ret+=t[k].mi[i]-now[i]; 68 if(now[i]>t[k].mx[i])ret+=now[i]-t[k].mx[i]; 69 } 70 return ret; 71 } 72 inline void query1(int k) 73 { 74 if(!k)return; 75 int dl=calc1(t[k].l),dr=calc1(t[k].r),d=dist(t[k],now); 76 if(d&&d<ans1)ans1=d; 77 if(dl<dr) 78 { 79 if(dl<ans1)query1(t[k].l); 80 if(dr<ans1)query1(t[k].r); 81 }else 82 { 83 if(dr<ans1)query1(t[k].r); 84 if(dl<ans1)query1(t[k].l); 85 } 86 } 87 inline int calc2(int k) 88 { 89 if(!k)return -inf; 90 ll ret=0; 91 for0(i,1)ret+=max(abs(t[k].mi[i]-now[i]),abs(t[k].mx[i]-now[i])); 92 return ret; 93 } 94 inline void query2(int k) 95 { 96 if(!k)return; 97 int dl=calc2(t[k].l),dr=calc2(t[k].r),d=dist(t[k],now); 98 if(d>ans2)ans2=d; 99 if(dl>dr) 100 { 101 if(dl>ans2)query2(t[k].l); 102 if(dr>ans2)query2(t[k].r); 103 }else 104 { 105 if(dr>ans2)query2(t[k].r); 106 if(dl>ans2)query2(t[k].l); 107 } 108 } 109 int main() 110 { 111 freopen("input.txt","r",stdin); 112 freopen("output.txt","w",stdout); 113 n=read(); 114 for1(i,n)p[i][0]=read(),p[i][1]=read(); 115 for0(i,1)t[0].mi[i]=inf,t[0].mx[i]=-inf; 116 int rt=build(1,n,0),ans=inf; 117 for1(i,n) 118 { 119 now=p[i]; 120 ans1=inf;ans2=-inf; 121 query1(rt); 122 query2(rt); 123 ans=min(ans,ans2-ans1); 124 } 125 cout<<ans<<endl; 126 return 0; 127 }