题意
一个k维空间,给出n个点的坐标,给出t个询问,每个询问给出一个点的坐标和一个m。对于每个询问找出跟这个点最接近的m个点
分析
kd树的模板题。
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <algorithm> 5 #include <queue> 6 7 using namespace std; 8 typedef long long LL; 9 const int maxn=50080; 10 const int K=5; 11 int num,nownum,m; 12 LL ans; 13 struct kdNode{ 14 LL x[K]; 15 int div; 16 bool lef; 17 }Ans[12]; 18 struct Node{ 19 kdNode a; 20 LL dis; 21 bool operator <(const Node &a)const{ 22 return dis<a.dis; 23 } 24 Node(){} 25 Node(kdNode &tmp,LL d){ 26 a=tmp; 27 dis=d; 28 } 29 }; 30 int cmpNo; 31 bool cmp(kdNode a,kdNode b){ 32 return a.x[cmpNo]<b.x[cmpNo]; 33 } 34 inline LL max(LL a,LL b){//why? 35 return a>b?a:b; 36 } 37 kdNode p[maxn],q; 38 LL dis(kdNode a,kdNode b,int k){ 39 LL res=0; 40 for(int i=0;i<k;i++){ 41 res+=(a.x[i]-b.x[i])*(a.x[i]-b.x[i]); 42 } 43 return res; 44 } 45 priority_queue<Node>qq; 46 void buildKD(int l,int r,kdNode* p,int d,int k){ 47 if(l>r)return ; 48 int m=(l+r)/2; 49 cmpNo=d; 50 nth_element(p+l,p+m,p+r+1,cmp); 51 p[m].div=d; 52 if(l==r){ 53 p[m].lef=1; 54 return ; 55 } 56 buildKD(l,m-1,p,(d+1)%k,k); 57 buildKD(m+1,r,p,(d+1)%k,k); 58 } 59 void findkd(int l,int r,kdNode& tar,kdNode* p,int k){ 60 if(l>r)return; 61 int m=(l+r)/2; 62 LL d=dis(p[m],tar,k); 63 if(p[m].lef){ 64 if(nownum<num){ 65 nownum++; 66 ans=max(ans,d); 67 qq.push(Node(p[m],d)); 68 } 69 else if(ans>d){ 70 qq.pop(); 71 qq.push(Node(p[m],d)); 72 ans=qq.top().dis; 73 } 74 return; 75 } 76 LL t=tar.x[p[m].div]-p[m].x[p[m].div]; 77 if(t>0){ 78 findkd(m+1,r,tar,p,k); 79 if(nownum<num){ 80 qq.push(Node(p[m],d)); 81 nownum++; 82 ans=qq.top().dis; 83 findkd(l,m-1,tar,p,k); 84 }else{ 85 if(ans>d){ 86 qq.pop(); 87 qq.push(Node(p[m],d)); 88 ans=qq.top().dis; 89 } 90 if(ans>t*t) 91 findkd(l,m-1,tar,p,k); 92 } 93 }else{ 94 findkd(l,m-1,tar,p,k); 95 if(nownum<num){ 96 qq.push(Node(p[m],d)); 97 nownum++; 98 ans=qq.top().dis; 99 findkd(m+1,r,tar,p,k); 100 }else{ 101 if(ans>d){ 102 qq.pop(); 103 qq.push(Node(p[m],d)); 104 ans=qq.top().dis; 105 } 106 if(ans>t*t){ 107 findkd(m+1,r,tar,p,k); 108 } 109 } 110 } 111 } 112 113 int n,k; 114 int main(){ 115 while(scanf("%d%d",&n,&k)==2){ 116 for(int i=0;i<n;i++){ 117 for(int j=0;j<k;j++){ 118 scanf("%lld",&p[i].x[j]); 119 } 120 p[i].lef=0; 121 } 122 buildKD(0,n-1,p,k-1,k); 123 int t; 124 scanf("%d",&t); 125 for(int i=1;i<=t;i++){ 126 ans=-1; 127 nownum=0; 128 for(int j=0;j<k;j++){ 129 scanf("%lld",&q.x[j]); 130 } 131 while(!qq.empty())qq.pop(); 132 scanf("%d",&num); 133 findkd(0,n-1,q,p,k); 134 for(int j=1;j<=num;j++){ 135 Ans[j]=qq.top().a; 136 qq.pop(); 137 } 138 printf("the closest %d points are:\n",num); 139 for(int j=num;j>=1;j--){ 140 for(int kk=0;kk<k;kk++){ 141 if(kk==0) 142 printf("%lld",Ans[j].x[kk]); 143 else 144 printf(" %lld",Ans[j].x[kk]); 145 } 146 printf("\n"); 147 } 148 } 149 } 150 return 0; 151 }
原文地址:https://www.cnblogs.com/LQLlulu/p/9894346.html
时间: 2024-11-08 05:46:03