BZOJ1941: [Sdoi2010]Hide and Seek

  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 }