hdu 2966 In case of failure(KD-tree)
题目链接：hdu 2966 In case of failure
题意：
给你n个点，让你输出每个点到最近点的欧式距离。
题解：
KD-树裸题，板子抄的鸟神的。
1 #include<bits/stdc++.h> 2 #define F(i,a,b) for(int i=(a);i<=(b);++i) 3 using namespace std; 4 using ll=long long; 5 6 namespace KD_Tree{ 7 const int N=1e5+7,DI=2; 8 struct Node{ 9 int p[DI],f,l,r,mx[DI],mi[DI]; 10 int operator[](const int&idx)const{return p[idx];} 11 void in(int idx,int *v){f=idx;F(i,0,DI-1)p[i]=v[i];} 12 void up(Node&a){ 13 F(i,0,DI-1){ 14 mi[i]=min(mi[i],a.mi[i]); 15 mx[i]=max(mx[i],a.mx[i]); 16 } 17 } 18 }T[N]; 19 int idx[N],cmpd,root; 20 21 bool cmp(const Node&a,const Node&b){return a[cmpd]<b[cmpd];} 22 void up(int x){ 23 if(T[x].l)T[x].up(T[T[x].l]); 24 if(T[x].r)T[x].up(T[T[x].r]); 25 } 26 int build(int l,int r,int d=0,int f=0) 27 { 28 int mid=l+r>>1; 29 cmpd=d%DI,nth_element(T+l,T+mid,T+r+1,cmp); 30 idx[T[mid].f]=mid,T[mid].f=f; 31 F(i,0,DI-1)T[mid].mi[i]=T[mid].mx[i]=T[f][i]; 32 T[mid].l=l!=mid?build(l,mid-1,d+1,mid):0; 33 T[mid].r=r!=mid?build(mid+1,r,d+1,mid):0; 34 return up(mid),mid; 35 } 36 ll dist(ll x,ll y=0){return x*x+y*y;} 37 ll query(ll x,ll y,ll&ans,int rt=root,int d=0) 38 { 39 ll tmp=dist(x-T[rt][0],y-T[rt][1]); 40 if(tmp)ans=min(ans,tmp); 41 if(T[rt].l&&T[rt].r) 42 { 43 bool is=!d?(x<=T[rt][0]):(y<=T[rt][1]); 44 ll dis=!d?dist(x-T[rt][0]):dist(y-T[rt][1]); 45 query(x,y,ans,is?T[rt].l:T[rt].r,!d); 46 if(dis<ans)query(x,y,ans,is?T[rt].r:T[rt].l,!d); 47 } 48 else if(T[rt].l)query(x,y,ans,T[rt].l,!d); 49 else if(T[rt].r)query(x,y,ans,T[rt].r,!d); 50 } 51 }; 52 using namespace KD_Tree; 53 int t,n; 54 55 int main(){ 56 scanf("%d",&t); 57 while(t--) 58 { 59 scanf("%d",&n); 60 F(i,1,n) 61 { 62 int x[2]; 63 scanf("%d%d",x,x+1); 64 T[i].in(i,x); 65 } 66 root=build(1,n); 67 F(i,1,n) 68 { 69 ll ans=1ll<<61; 70 query(T[idx[i]][0],T[idx[i]][1],ans); 71 printf("%lld\n",ans); 72 } 73 } 74 return 0; 75 }
