K-D Tree题目泛做（CXJ第二轮）

题目1： BZOJ 2716

题目大意：给出N个二维平面上的点，M个操作，分为插入一个新点和询问到一个点最近点的Manhatan距离是多少。

算法讨论：

K-D Tree 裸题，有插入操作。

 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>

 using namespace std;

 const int inf = 1e9;
 const int K = ;
 const int N =  + ;

 inline int read() {
   int x = ;
   char ch = getchar();

   while(ch < '' || ch > '') ch = getchar();
   while(ch <= '' && ch >= '') {
     x = x *  + ch - '';
     ch = getchar();
   }
   return x;
 }

 int n, m, root, D, ans;

 struct Node {
   int d[K], mn[K], mx[K], l, r, v, sum;

   int & operator [] (int x) {
     return d[x];
   }
   Node(int x = , int y = ) {
     l = ; r = ; d[] = x; d[] = y;
   }
   friend bool operator < (Node a, Node b) {
     return a[D] < b[D];
   }
 }p[N], T[N<<], tmp;

 void pushup(int k) {
   Node l = T[T[k].l], r = T[T[k].r];

   for(int i = ; i < K; ++ i) {
     T[k].mn[i] = T[k].mx[i] = T[k][i];
     if(T[k].l) {
       T[k].mn[i] = min(T[k].mn[i], l.mn[i]);
       T[k].mx[i] = max(T[k].mx[i], l.mx[i]);
     }
     if(T[k].r) {
       T[k].mx[i] = max(T[k].mx[i], r.mx[i]);
       T[k].mn[i] = min(T[k].mn[i], r.mn[i]);
     }
   }
   //  T[k].sum = T[k].v;
   //  if(T[k].l) T[k].sum += l.sum;
   //  if(T[k].r) T[k].sum += r.sum;
 }

 int build(int l, int r, int nd) {
   int mid = (l + r) >> ;

   D = nd;
   nth_element(p + l, p + mid, p + r + );
   T[mid] = p[mid];
   T[mid].l = T[mid].r = ;
   for(int i = ; i < K; ++ i)
     T[mid].mx[i] = T[mid].mn[i] = T[mid][i];
   if(l < mid)
     T[mid].l = build(l, mid - , (nd + ) % K);
   if(r > mid)
     T[mid].r = build(mid + , r, (nd + ) % K);
   pushup(mid);
   return mid;
 }

 void insert(int k, int nd) {
   if(tmp[nd] >= T[k][nd]) {
     if(T[k].r) insert(T[k].r, (nd + ) % K);
     else {
       T[k].r = ++ n;
       T[n] = tmp;
       for(int i = ; i < K; ++ i)
         T[n].mn[i] = T[n].mx[i] = T[n][i];
     }
   }
   else {
     if(T[k].l) insert(T[k].l, (nd + ) % K);
     else {
       T[k].l = ++ n;
       T[n] = tmp;
       for(int i = ; i < K; ++ i)
         T[n].mn[i] = T[n].mx[i] = T[n][i];
     }
   }
   pushup(k);
 }

 int getkdis(Node a, Node b) {
   int res = ;

   for(int i = ; i < K; ++ i)
     res += abs(a[i] - b[i]);
   return res;
 }

 int inandout(int k, Node a) {
   int res = ;

   for(int i = ; i < K; ++ i)
     res += max(, T[k].mn[i] - a[i]);
   for(int i = ; i < K; ++ i)
     res += max(, a[i] - T[k].mx[i]);
   return res;
 }

 void query(int k, int nd) {
   int d, dl = inf, dr = inf;

   d = getkdis(T[k], tmp);
   if(d) ans = min(ans, d);
   if(T[k].l) dl = inandout(T[k].l, tmp);
   if(T[k].r) dr = inandout(T[k].r, tmp);
   if(dl < dr) {
     if(dl < ans) query(T[k].l, (nd + ) % K);
     if(dr < ans) query(T[k].r, (nd + ) % K);
   }
   else {
     if(dr < ans) query(T[k].r, (nd + ) % K);
     if(dl < ans) query(T[k].l, (nd + ) % K);
   }
 }

 int query(Node a) {
   tmp = a; ans = inf;
   query(root, );
   return ans;
 }

 void insert(Node a) {
   tmp = a; insert(root, );
 }
 #define ONLINE_JUDGE
 int main() {
 #ifndef ONLINE_JUDGE
   freopen("1.in", "r", stdin);
   freopen("1.out", "w", stdout);
 #endif

   int type, x, y;

   n = read(); m = read();
   for(int i = ; i <= n; ++ i)
     p[i][] = read(), p[i][] = read();
   root = build(, n, );//忘记写root等于了。
   for(int i = ; i <= m; ++ i) {
     type = read(); x = read(); y = read();
     if(type == ) insert(Node(x, y));
     else printf("%d\n", query(Node(x, y)));
   }

 #ifndef ONLINE_JUDGE
   fclose(stdin); fclose(stdout);
 #endif
   return ;
 }

BZOJ 2716

题目2： BZOJ 1941

题目大意：给出N个点，求对于每个点说，Manhantan距离最远点与最近点的差值最小是多少。

算法讨论：

K-D Tree裸题，注意最近点不能是自己。

 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>

 using namespace std;

 const int N =  + ;
 const int inf = 1e9;
 const int K = ;

 inline int read() {
   int x = ;
   char ch = getchar();

   while(ch < '' || ch > '') ch = getchar();
   while(ch <= '' && ch >= '') {
     x = x *  + ch - '';
     ch = getchar();
   }
   return x;
 }

 int n, root, amax, amin, D;

 struct Node {
   int d[K], mn[K], mx[K], l, r;

   int & operator [] (int x) {
     return d[x];
   }
   Node (int x = , int y = ) {
     l = ; r = ; d[] = x; d[] = y;
   }
   friend bool operator < (Node a, Node b) {
     return a[D] < b[D];
   }
 }p[N], T[N<<], tmp;

 void pushup(int k) {
   Node l = T[T[k].l], r = T[T[k].r];

   for(int i = ; i < K; ++ i) {
     T[k].mn[i] = T[k].mx[i] = T[k][i];
     if(T[k].l) {
       T[k].mx[i] = max(T[k].mx[i], l.mx[i]);
       T[k].mn[i] = min(T[k].mn[i], l.mn[i]);
     }
     if(T[k].r) {
       T[k].mx[i] = max(T[k].mx[i], r.mx[i]);
       T[k].mn[i] = min(T[k].mn[i], r.mn[i]);
     }
   }
 }

 int build(int l, int r, int nd) {
   int mid = (l + r) >> ;

   D = nd;
   nth_element(p + l, p + mid, p + r + );
   T[mid] = p[mid];
   T[mid].l = T[mid].r = ;
   for(int i = ; i < K; ++ i)
     T[mid].mn[i] = T[mid].mx[i] = T[mid][i];
   if(l < mid) T[mid].l = build(l, mid - , (D + ) % K);
   if(r > mid) T[mid].r = build(mid + , r, (D + ) % K);
   pushup(mid);
   return mid;
 }

 void insert(int k, int nd) {
   if(tmp[nd] >= T[k][nd]) {
     if(T[k].r) insert(T[k].r, (nd + ) % K);
     else {
       T[k].r = ++ n;
       T[n] = tmp;
       for(int i = ; i < K; ++ i)
         T[n].mx[i] = T[n].mn[i] = T[n][i];
     }
   }
   else {
     if(T[k].l) insert(T[k].l, (nd + ) % K);
     else {
       T[k].l = ++ n;
       T[n] = tmp;
       for(int i = ; i < K; ++ i)
         T[n].mx[i] = T[n].mn[i] = T[n][i];
     }
   }
   pushup(k);
 }

 int getkdis(Node a, Node b) {
   int res = ;

   for(int i = ; i < K; ++ i)
     res += abs(a[i] - b[i]);
   return res;
 }

 int outandin(int k, Node q) {
   int res = ;

   for(int i = ; i < K; ++ i)
     res += max(, T[k].mn[i] - q[i]);
   for(int i = ; i < K; ++ i)
     res += max(, q[i] - T[k].mx[i]);
   return res;
 }

 int outandinmaxx(int k, Node q) {
   int res = ;

   for(int i = ; i < K; ++ i)
     res += max(abs(T[k].mn[i] - q[i]), abs(T[k].mx[i] - q[i]));
   return res;
 }

 void query_maxx(int k, int nd) {
   int d, dl = -inf, dr = -inf;

   d = getkdis(T[k], tmp);
   amax = max(d, amax);
   if(T[k].l) dl = outandinmaxx(T[k].l, tmp);
   if(T[k].r) dr = outandinmaxx(T[k].r, tmp);
   if(dl > dr) {
     if(dl > amax) query_maxx(T[k].l, (nd + ) % K);
     if(dr > amax) query_maxx(T[k].r, (nd + ) % K);
   }
   else {
     if(dr > amax) query_maxx(T[k].r, (nd + ) % K);
     if(dl > amax) query_maxx(T[k].l, (nd + ) % K);
   }
 }

 void query_minn(int k, int nd) {
   int d, dl = inf, dr = inf;

   d = getkdis(T[k], tmp);
   if(d) amin = min(d, amin);
   if(T[k].l) dl = outandin(T[k].l, tmp);
   if(T[k].r) dr = outandin(T[k].r, tmp);
   if(dl < dr) {
     if(dl < amin) query_minn(T[k].l, (nd + ) % K);
     if(dr < amin) query_minn(T[k].r, (nd + ) % K);
   }
   else {
     if(dr < amin) query_minn(T[k].r, (nd + ) % K);
     if(dl < amin) query_minn(T[k].l, (nd + ) % K);
   }
 }

 void qmax(int l) {
   amax = -inf; tmp = p[l];
   query_maxx(root, );

 }

 void qmin(int l) {
   amin = inf; tmp = p[l];
   query_minn(root, );
 }

 int main() {
   int outans = inf;

   n = read();
   for(int i = ; i <= n; ++ i) {
     p[i][] = read(); p[i][] = read();
   }
   root = build(, n, );
   for(int i = ; i <= n; ++ i) {
     qmax(i); qmin(i);
     outans = min(outans, amax - amin);
   }
   printf("%d\n", outans);
   return ;
 }

BZOJ 1941

题目3： BZOJ4520 &&　CQOI 2016 K远点对查询

题目大意：

给出n个二维平面上的点，求第k远的点对距离是多少。(欧几里德距离的平方)

算法讨论：

1、为了防止重复，小根堆里面保存2*K个元素。

2、编程习惯一定要好。同类型比较，减少强制类型转制。在查询欧几里德距离的时候，query中的维度参数是没有用的。

果断要去掉。否则就是TLE和AC的区别。

要区分好查询欧几里德距离和曼哈顿距离时两者的区别。

代码：

#include <cstdlib>
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <queue>
#include <vector>

using namespace std;
typedef long long ll;
const int N = 100000 + 5;
const int K = 2;
const ll inf = 10000000000000LL;
inline int read() {
  int x = 0; char c = getchar();
  while(!isdigit(c)) c = getchar();
  while(isdigit(c)) {
	x = x * 10 + c - '0';
	c = getchar();
  }
  return x;
}
int buf[20];
inline void output(ll x) {
  int p = 0; buf[0] = 0;
  if(!x) p ++;
  else {
	while(x) {
	  buf[p ++] = x % 10;
	  x /= 10;
	}
  }
  for(int i = p - 1; i >= 0; -- i)
	putchar(buf[i] + '0');
}

int root, n, kk, D;
priority_queue <ll, vector<ll>, greater<ll> > ans;

struct Node {
  int l, r;
  ll d[K], mn[K], mx[K];
  Node(ll x = 0, ll y = 0) {
	l = r = 0; d[0] = x; d[1] = y;
  }
  ll & operator [] (int x) { return d[x];}
  friend bool operator < (Node a, Node b) {
	return a[D] < b[D];
  }
}p[N], T[N << 1], tmp;

void pushup(int k) {
  Node l = T[T[k].l], r = T[T[k].r];
  for(int i = 0; i < K; ++ i) {
	T[k].mx[i] = T[k].mn[i] = T[k][i];
	if(T[k].l) {
	  T[k].mx[i] = max(T[k].mx[i], l.mx[i]);
	  T[k].mn[i] = min(T[k].mn[i], l.mn[i]);
	}
	if(T[k].r) {
	  T[k].mx[i] = max(T[k].mx[i], r.mx[i]);
	  T[k].mn[i] = min(T[k].mn[i], r.mn[i]);
	}
  }
}

int build(int l, int r, int nd) {
  int mid = (l + r) >> 1;
  D = nd;
  nth_element(p + l, p + mid, p + r + 1);
  T[mid] = p[mid];
  T[mid].l = T[mid].r = 0;
  for(int i = 0; i < K; ++ i)
	T[mid].mx[i] = T[mid].mn[i] = T[mid][i];
  if(l < mid) T[mid].l = build(l, mid - 1, (nd + 1) % K);
  if(r > mid) T[mid].r = build(mid + 1, r, (nd + 1) % K);
  pushup(mid);
  return mid;
}

ll geteulerdis(Node a, Node b) {//竭诚为欧几里德距离服务
  ll res = 0;
  for(int i = 0; i < K; ++ i)
	res += (a[i] - b[i]) * (a[i] - b[i]);
  return res;
}

ll outandineuler(Node a) {
  ll L = 0;
  L = max(L, geteulerdis(tmp, Node(a.mx[0], a.mn[1])));
  L = max(L, geteulerdis(tmp, Node(a.mx[0], a.mx[1])));
  L = max(L, geteulerdis(tmp, Node(a.mn[0], a.mn[1])));
  L = max(L, geteulerdis(tmp, Node(a.mn[0], a.mx[1])));
  return L;
}

void query(int k) {
  ll d, dl = -inf, dr = -inf;
  d = geteulerdis(T[k], tmp);
  if(d > ans.top()) {
	ans.pop(); ans.push(d);
  }
  if(T[k].l) dl = outandineuler(T[T[k].l]);
  if(T[k].r) dr = outandineuler(T[T[k].r]);
  if(dl > dr) {
	if(dl > ans.top()) query(T[k].l);
	if(dr > ans.top()) query(T[k].r);
  }
  else {
	if(dr > ans.top()) query(T[k].r);
	if(dl > ans.top()) query(T[k].l);
  }
}

void Q(int i) {
  tmp = p[i];
  query(root);
}

int main() {
  n = read(); kk = read();
  for(int i = 1; i <= 2 * kk; ++ i) ans.push(0);
  for(int i = 1; i <= n; ++ i) {
	p[i][0] = read(); p[i][1] = read();
  }
  root = build(1, n, 0);
  for(int i = 1; i <= n; ++ i) Q(i);
  output(ans.top());
  return 0;
}