【BZOJ 1941】[Sdoi2010]Hide and Seek kd-tree

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裸的kd-tree

1. nod[0]初始化

2. 类似于splay的建树方法，维护n个节点，每一个节点管辖一定区域

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<cstdlib>#include<cmath>#define maxn 500021#define inf 0x3fffffff#define ls(u) nod[u].l#define rs(u) nod[u].r#define rep(i) for(int i=0;i<2;i++)using namespace std;int n,ans1,ans2,pos,rt,dx,dy;struct node{int d[2],Max[2],Min[2],l,r;bool operator<(const node& b)const{return d[pos]==b.d[pos] ? d[pos^1]<b.d[pos^1] : d[pos]<b.d[pos];}}nod[maxn];void push_up(int u){rep(i){nod[u].Min[i]=min(nod[u].d[i],min(nod[ls(u)].Min[i],nod[rs(u)].Min[i]));nod[u].Max[i]=max(nod[u].d[i],max(nod[ls(u)].Max[i],nod[rs(u)].Max[i]));}}int build(int l,int r,int k){if(l>r)return 0;pos=k;int mid=l+r>>1;nth_element(nod+1+l,nod+1+mid,nod+1+r);nod[mid].Max[0]=nod[mid].Min[0]=nod[mid].d[0];nod[mid].Max[1]=nod[mid].Min[1]=nod[mid].d[1];if(l==r)return l;nod[mid].l=build(l,mid-1,k^1);nod[mid].r=build(mid+1,r,k^1);push_up(mid);return mid;}int Q(int u){return abs(nod[u].d[0]-dx)+abs(nod[u].d[1]-dy);}int guess(int u){if(!u)return inf;int ans=0;ans+=max(0,dx-nod[u].Max[0])+max(0,nod[u].Min[0]-dx);ans+=max(0,dy-nod[u].Max[1])+max(0,nod[u].Min[1]-dy);return ans;}void query(int u){if(!u)return;int a0=Q(u),a1=guess(ls(u)),a2=guess(rs(u));if(a0)ans1=min(ans1,a0);if(a1<a2){if(ans1>a1)query(ls(u));if(ans1>a2)query(rs(u));}else{if(ans1>a2)query(rs(u));if(ans1>a1)query(ls(u));}}int guess2(int u){if(!u)return -inf;return max(dx-nod[u].Min[0],nod[u].Max[0]-dx)+max(dy-nod[u].Min[1],nod[u].Max[1]-dy);}void query2(int u){if(!u)return;int a0=Q(u),a1=guess2(ls(u)),a2=guess2(rs(u));ans2=max(ans2,a0);if(a1>a2){if(ans2<a1)query2(ls(u));if(ans2<a2)query2(rs(u));}else{if(ans2<a2)query2(rs(u));if(ans2<a1)query2(ls(u));}}int main(){nod[0].Min[0]=nod[0].Min[1]=inf;nod[0].Max[0]=nod[0].Max[1]=-inf;scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d%d",&nod[i].d[0],&nod[i].d[1]);rt=build(1,n,0);int a1=inf;for(int i=1;i<=n;i++){dx=nod[i].d[0],dy=nod[i].d[1];ans1=inf,ans2=-inf;query(rt),query2(rt);a1=min(a1,ans2-ans1);}printf("%d",a1);return 0;}

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