0.0找第k大……
裸KD-Tree……跟之前那道找最近的k个点大同小异
一开始理解错：第K大是第K远……不是第K近……（Tunix你个sb
感觉容易出错的是0号点= =边界情况需要仔细处理……根据题意而定的，比如这题就必须将0号点的距离设置成最近……比如-2……（因为我一开始向堆里加的占位点的距离是-1
1 /************************************************************** 2 Problem: 2626 3 User: Tunix 4 Language: C++ 5 Result: Accepted 6 Time:12956 ms 7 Memory:4796 kb 8 ****************************************************************/ 9 10 //BZOJ 2626 11 #include<queue> 12 #include<cstdio> 13 #include<cstring> 14 #include<cstdlib> 15 #include<iostream> 16 #include<algorithm> 17 #define rep(i,n) for(int i=0;i<n;++i) 18 #define F(i,j,n) for(int i=j;i<=n;++i) 19 #define D(i,j,n) for(int i=j;i>=n;--i) 20 #define pb push_back 21 #define sqr(x) (x)*(x) 22 using namespace std; 23 typedef long long LL; 24 inline int getint(){ 25 int r=1,v=0; char ch=getchar(); 26 for(;!isdigit(ch);ch=getchar()) if (ch=='-') r=-1; 27 for(; isdigit(ch);ch=getchar()) v=v*10-'0'+ch; 28 return r*v; 29 } 30 const int N=100010,INF=1e9; 31 /*******************template********************/ 32 int n,m,root,D; 33 struct Poi{ 34 int d[2],mn[2],mx[2],l,r,num; 35 int& operator [] (int x){return d[x];} 36 void read(){d[0]=getint();d[1]=getint();} 37 }t[N],now; 38 39 bool operator < (Poi a,Poi b) {return a[D]<b[D];} 40 struct node{ 41 LL dis;int num; 42 node(LL d=0,int n=0):dis(d),num(n){} 43 }; 44 bool operator < (const node &a,const node &b){ 45 return a.dis>b.dis || (a.dis==b.dis && a.num<b.num); 46 } 47 #define L t[o].l 48 #define R t[o].r 49 #define mid (l+r>>1) 50 void Push_up(int o){ 51 F(i,0,1){ 52 t[o].mn[i]=min(t[o].mn[i],min(t[L].mn[i],t[R].mn[i])); 53 t[o].mx[i]=max(t[o].mx[i],max(t[L].mx[i],t[R].mx[i])); 54 } 55 } 56 int build(int l,int r,int dir){ 57 D=dir; 58 nth_element(t+l,t+mid,t+r+1); 59 int o=mid; 60 F(i,0,1) t[o].mn[i]=t[o].mx[i]=t[o][i]; 61 if (l<mid) L=build(l,mid-1,dir^1); 62 if (r>mid) R=build(mid+1,r,dir^1); 63 Push_up(o); 64 return o; 65 } 66 67 inline LL dis(Poi a,Poi b){return (LL)sqr(a[0]-b[0])+(LL)sqr(a[1]-b[1]);} 68 inline LL getdis(int o){ 69 if (!o) return -2; 70 LL ans=0; 71 F(i,0,1) 72 ans+=max((LL)sqr(t[o].mx[i]-now[i]),(LL)sqr(t[o].mn[i]-now[i])); 73 return ans; 74 } 75 priority_queue<node>Q; 76 77 void query(int o){ 78 if (!o) return; 79 LL dl=getdis(L),dr=getdis(R),d0=dis(t[o],now); 80 if (d0>Q.top().dis || (d0==Q.top().dis && t[o].num<Q.top().num)){ 81 Q.pop(); Q.push(node(d0,t[o].num)); 82 } 83 if (dl>dr){ 84 if (dl>=Q.top().dis) query(L); 85 if (dr>=Q.top().dis) query(R); 86 }else{ 87 if (dr>=Q.top().dis) query(R); 88 if (dl>=Q.top().dis) query(L); 89 } 90 } 91 92 int main(){ 93 #ifndef ONLINE_JUDGE 94 freopen("2626.in","r",stdin); 95 freopen("2626.out","w",stdout); 96 #endif 97 F(i,0,1) t[0].mn[i]=INF,t[0].mx[i]=-INF; 98 n=getint(); 99 F(i,1,n) t[i].read(),t[i].num=i; 100 root=build(1,n,1); 101 int T=getint(); 102 while(T--){ 103 now.read(); int k=getint(); 104 // printf("%d %d %d\n",now[0],now[1],k); 105 while(!Q.empty()) Q.pop(); 106 F(i,1,k) Q.push(node(-1,0)); 107 query(root); 108 printf("%d\n",Q.top().num); 109 // printf("Q.size()=%d\n",Q.size()); 110 // F(i,1,k) 111 // printf("%d -- %lld %lld\n",i,Q.top().dis,Q.top().num),Q.pop(); 112 // puts(""); 113 } 114 return 0; 115 }