记录编号	425174	评测结果	AAAAAAAAAAAAAAAAAAAAAAAAA
题目名称	[Keller战纪·正传·妖姬篇][HZOI 2015]Keller非.a.t.e	最终得分	100
用户昵称	rewine	是否通过	通过
代码语言	C++	运行时间	29.585 s
提交时间	2017-07-14 19:53:57	内存使用	20.15 MiB

//http://blog.sina.cn/dpool/blog/s/blog_6f611c300101bysf.html
#include <cstdio>
#include <iostream>
#include <queue>
#include <algorithm>

using namespace std;

typedef long long LL;
typedef long double LB;


void read(LL &x) {static char c; static bool flag;for (flag = 0; (c=getchar())<'0'||c>'9'; flag |= c=='-');for (x=c-'0'; (c=getchar())>='0'&&c<='9'; x=x*10+c-'0');if(flag) x = -x;}

inline LL F(LL x) {return x*x*x*x;}
#define N 200000 

LL D;

struct Data {
  LL t,v[5];
  int flag;
  void in() {
    read(t);
    for (int i = 0; i < t; i++) read(v[i]);
    for (int i = t; i < 5; i++) v[i] = 0;
  }
  bool operator < (const Data &lhs) const {
  	return v[D] < lhs.v[D];   
  }
}tt[N],t[N],q;

priority_queue<LB> ans;

LL C,n,Q;

LB Dis(const Data &a,const Data &b) {
   LB t = 0;
   for (int i = 0; i < 5; i++) 
     t += F(a.v[i]-b.v[i]);
   return t;
}

void Push(LB v) {
   ans.push(v);
   if(ans.size()>C) ans.pop();
}

void Out() {
	static vector<LB> t;
	t.clear();
	while(ans.size()) {
	   t.push_back(ans.top()); ans.pop();
	}
	for (int i = t.size()-1; i >= 0; i--)
  	  printf("%lld\n",(LL)(t[i]+0.5));
}
///////////

#define L (o<<1) 
#define R (o<<1|1)
int rt;
void build(int l,int r,int o,int cas) {
    if(l>r) return; D = cas;
    t[o].flag = 1; t[L].flag = t[R].flag = -1;
    int mid = (l+r)>>1;
    nth_element(tt+l,tt+mid,tt+r+1);
    t[o] = tt[mid];
    cas = (cas+1)%5;
    build(l,mid-1,L,cas); build(mid+1,r,R,cas);
}

void query(int o,int cas) {
    if(t[o].flag == -1) return;
    int x = L,y = R;
	if(q.v[cas] > t[o].v[cas]) swap(x,y);
	query(x,(cas+1)%5);
	if(ans.size() < C) {
		Push(Dis(t[o],q));
		query(y,(cas+1)%5);
	} else {
	    Push(Dis(t[o],q));
	    if(F(t[o].v[cas]-q.v[cas]) <= ans.top()) query(y,(cas+1)%5);
	}
}


int main() {
  freopen("Keller_Deal.in","r",stdin);freopen("Keller_Deal.out","w",stdout);
  read(n); read(Q);
  for (int i = 1; i <= n; i++) tt[i].in();
  build(1,n,1,0);
  for (int i = 1; i <= Q; i++) {
    read(q.t); read(C);
    for (int i = 0; i < q.t; i++) read(q.v[i]);
    for (int i = q.t; i < 5; i++) q.v[i] = 0;
    printf("the closest %lld strings' distance are:\n",C);
    query(1,0);
    Out(); 
  }
  return 0;
}