【BZOJ】【3053】The Closest M Points
KD-Tree
题目大意：K维空间内，与给定点欧几里得距离最近的 m 个点。
KD树啊……还能怎样啊……然而扩展到k维其实并没多么复杂？除了我已经脑补不出建树过程……不过代码好像变化不大>_>
然而我WA了。。。为什么呢。。。我也不知道……
一开始我的Push_up是这么写的：
inline void Push_up(int o){ rep(i,k){ if (L) t[o].mn[i]=min(t[o].mn[i],t[L].mn[i]),t[o].mx[i]=max(t[o].mx[i],t[L].mx[i]); if (R) t[o].mx[i]=min(t[o].mn[i],t[R].mn[i]),t[o].mx[i]=max(t[o].mx[i],t[R].mx[i]); } }
就是如果没有右儿子，就不用它更新了……
然而我改成zyf的这样：（t[0]初始化一下，mn都置为INF，mx都置为-INF）
inline void Push_up(int o){ rep(i,k){ t[o].mn[i]=min(t[o].mn[i],min(t[L].mn[i],t[R].mn[i])); t[o].mx[i]=max(t[o].mx[i],max(t[L].mx[i],t[R].mx[i])); } }
就过了………………
P.S.多组数据的题目，KD-Tree一定要记得清空 l 和 r ……要不然叶子节点会有一些奇怪的问题……
获得称号：
1 /************************************************************** 2 Problem: 3053 3 User: Tunix 4 Language: C++ 5 Result: Accepted 6 Time:1364 ms 7 Memory:9088 kb 8 ****************************************************************/ 9 10 //BZOJ 3053 11 #include<queue> 12 #include<cmath> 13 #include<cstdio> 14 #include<cstring> 15 #include<cstdlib> 16 #include<iostream> 17 #include<algorithm> 18 #define rep(i,n) for(int i=0;i<n;++i) 19 #define F(i,j,n) for(int i=j;i<=n;++i) 20 #define D(i,j,n) for(int i=j;i>=n;--i) 21 #define pb push_back 22 #define sqr(x) ((x)*(x)) 23 using namespace std; 24 typedef long long LL; 25 inline int getint(){ 26 int r=1,v=0; char ch=getchar(); 27 for(;!isdigit(ch);ch=getchar()) if (ch=='-') r=-1; 28 for(; isdigit(ch);ch=getchar()) v=v*10-'0'+ch; 29 return r*v; 30 } 31 const int N=100010,INF=1e9; 32 /*******************template********************/ 33 34 int n,k,D,root,ans[15]; 35 struct node{ 36 int d[6],mn[6],mx[6],l,r; 37 int& operator [] (int x){return d[x];} 38 void read(){rep(i,k) d[i]=getint();} 39 }t[N],tmp; 40 bool operator < (node a,node b){return a[D]<b[D];} 41 #define L t[o].l 42 #define R t[o].r 43 #define mid (l+r>>1) 44 inline void Push_up(int o){ 45 rep(i,k){ 46 t[o].mn[i]=min(t[o].mn[i],min(t[L].mn[i],t[R].mn[i])); 47 t[o].mx[i]=max(t[o].mx[i],max(t[L].mx[i],t[R].mx[i])); 48 } 49 } 50 int build(int l,int r,int dir){ 51 D=dir; 52 nth_element(t+l,t+mid,t+r+1); 53 rep(i,k) t[mid].mn[i]=t[mid].mx[i]=t[mid][i]; 54 t[mid].l=l<mid ? build(l,mid-1,(dir+1)%k) : 0; 55 t[mid].r=mid<r ? build(mid+1,r,(dir+1)%k) : 0; 56 Push_up(mid); 57 return mid; 58 } 59 inline int getdis(int o){ 60 if (!o) return INF; 61 int ans=0; 62 rep(i,k) if (tmp[i]<t[o].mn[i]) ans+=sqr(t[o].mn[i]-tmp[i]); 63 rep(i,k) if (tmp[i]>t[o].mx[i]) ans+=sqr(tmp[i]-t[o].mx[i]); 64 return ans; 65 } 66 inline int dis(node a,node b){ 67 int ans=0; 68 rep(i,k) ans+=sqr(a[i]-b[i]); 69 return ans; 70 } 71 typedef pair<int,int> pii; 72 priority_queue<pii>Q; 73 #define mp make_pair 74 #define X first 75 #define Y second 76 void query(int o){ 77 int dl=getdis(L),dr=getdis(R),d0=dis(t[o],tmp); 78 if (d0<Q.top().X){Q.pop(); Q.push(mp(d0,o));} 79 if (dl<dr){ 80 if (dl<Q.top().X) query(L); 81 if (dr<Q.top().X) query(R); 82 }else{ 83 if (dr<Q.top().X) query(R); 84 if (dl<Q.top().X) query(L); 85 } 86 } 87 int main(){ 88 #ifndef ONLINE_JUDGE 89 freopen("3053.in","r",stdin); 90 freopen("3053.out","w",stdout); 91 #endif 92 rep(i,5) t[0].mn[i]=INF,t[0].mx[i]=-INF; 93 while(scanf("%d%d",&n,&k)!=EOF){ 94 F(i,1,n) rep(j,k) t[i][j]=getint(); 95 root=build(1,n,0); 96 int T=getint(); 97 while(T--){ 98 rep(j,k) tmp[j]=getint(); 99 n=getint(); 100 printf("the closest %d points are:\n",n); 101 F(i,1,n) Q.push(mp(INF,0)); 102 query(root); 103 D(i,n,1) ans[i]=Q.top().Y,Q.pop(); 104 F(i,1,n) rep(j,k) 105 printf("%d%c",t[ans[i]][j],j!=k-1?' ':'\n'); 106 } 107 } 108 return 0; 109 }
3053: The Closest M Points
Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 383 Solved: 147
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Description
The
course of Software Design and Development Practice is objectionable.
ZLC is facing a serious problem .There are many points in K-dimensional
space .Given a point. ZLC need to find out the closest m points.
Euclidean distance is used as the distance metric between two points.
The Euclidean distance between points p and q is the length of the line
segment connecting them.In Cartesian coordinates, if p = (p1, p2,...,
pn) and q = (q1, q2,..., qn) are two points in Euclidean n-space, then
the distance from p to q, or from q to p is given by:
D(p,q)=D(q,p)=sqrt((q1-p1)^2+(q2-p2)^2+(q3-p3)^2…+(qn-pn)^2
Can you help him solve this problem?
软工学院的课程很讨厌！ZLC同志遇到了一个头疼的问题：在K维空间里面有许多的点，对于某些给定的点，ZLC需要找到和它最近的m个点。
（这里的距离指的是欧几里得距离：D(p, q) = D(q, p) = sqrt((q1 - p1) ^ 2 + (q2 - p2) ^ 2 + (q3 - p3) ^ 2 + ... + (qn - pn) ^ 2)
ZLC要去打Dota，所以就麻烦你帮忙解决一下了……
【Input】
第一行，两个非负整数：点数n(1 <= n <= 50000)，和维度数k(1 <= k <= 5)。
接下来的n行，每行k个整数，代表一个点的坐标。
接下来一个正整数：给定的询问数量t(1 <= t <= 10000)
下面2*t行：
第一行，k个整数：给定点的坐标
第二行：查询最近的m个点(1 <= m <= 10)
所有坐标的绝对值不超过10000。
有多组数据！
【Output】
对于每个询问，输出m+1行：
第一行："the closest m points are:" m为查询中的m
接下来m行每行代表一个点，按照从近到远排序。
保证方案唯一，下面这种情况不会出现：
2 2
1 1
3 3
1
2 2
1
Input
In the
first line of the text file .there are two non-negative integers n and
K. They denote respectively: the number of points, 1 <= n <=
50000, and the number of Dimensions,1 <= K <= 5. In each of the
following n lines there is written k integers, representing the
coordinates of a point. This followed by a line with one positive
integer t, representing the number of queries,1 <= t <=10000.each
query contains two lines. The k integers in the first line represent the
given point. In the second line, there is one integer m, the number of
closest points you should find,1 <= m <=10. The absolute value of
all the coordinates will not be more than 10000.
There are multiple test cases. Process to end of file.
Output
For each query, output m+1 lines:
The first line saying :”the closest m points are:” where m is the number of the points.
The following m lines representing m points ,in accordance with the order from near to far
It is guaranteed that the answer can only be formed in one ways. The
distances from the given point to all the nearest m+1 points are
different. That means input like this:
2 2
1 1
3 3
1
2 2
1
will not exist.
Sample Input
1 1
1 3
3 4
2
2 3
2
2 3
1
Sample Output
1 3
3 4
the closest 1 points are:
1 3
