AOJ DSL_2_C Range Search (kD Tree)

Range Search (kD Tree)

The range search problem consists of a set of attributed records S to determine which records from Sintersect with a given range.

For n points on a plane, report a set of points which are within in a given range. Note that you do not need to consider insert and delete operations for the set.

Input

n
x0 y0
x1 y1
:
xn竏�1 yn竏�1
q
sx0 tx0 sy0 ty0
sx1 tx1 sy1 ty1
:
sxq竏�1 txq竏�1 syq竏�1 tyq竏�1

The first integer n is the number of points. In the following n lines, the coordinate of the i-th point is given by two integers xi and yi.

The next integer q is the number of queries. In the following q lines, each query is given by four integers,sxi, txi, syi, tyi.

Output

For each query, report IDs of points such that sxi ≤ x ≤ txi and syi ≤ y ≤ tyi. The IDs should be reported in ascending order. Print an ID in a line, and print a blank line at the end of output for the each query.

Constraints

  • 0 ≤ n ≤ 500,000
  • 0 ≤ q ≤ 20,000
  • -1,000,000,000 ≤ x, y, sx, tx, sy, ty ≤ 1,000,000,000
  • sx ≤ tx
  • sy ≤ ty
  • For each query, the number of points which are within the range is less than or equal to 100.

Sample Input 1

6
2 1
2 2
4 2
6 2
3 3
5 4
2
2 4 0 4
4 10 2 5

Sample Output 1

0
1
2
4

2
3
5
 
 
虽然说是KD-Tree模板题，但是貌似怎么做都可以；当然拿来练习KD-Tree还是十分不错的。于是压了压常数就在AOJ上打榜了，谁叫JP人总喜欢用cin和STL呢，( ﹁ ﹁ ) ~→
 
  1 #include <cstdio>
  2 #include <iostream>
  3 #include <algorithm>
  4 using namespace std;
  5 
  6 inline int next(void) {
  7     register int ret = 0;
  8     register int neg = false;
  9     register int bit = getchar();
 10     for (; bit < '0'; bit = getchar())
 11         if (bit == '-')neg ^= true;
 12     for (; bit >= '0'; bit = getchar())
 13         ret = ret * 10 + bit - '0';
 14     return neg ? -ret : ret;
 15 }
 16 
 17 int n, m;
 18 
 19 int root;
 20 int cmpk;
 21 
 22 struct node {
 23     int id;
 24     int pos[2];
 25     int son[2];
 26     int min[2];
 27     int max[2];
 28     inline friend bool operator < 
 29         (const node &a, const node &b) {
 30             return a.pos[cmpk] < b.pos[cmpk];
 31     }
 32 }tree[500005];
 33 
 34 inline void update(int t) {
 35     for (int k = 0; k < 2; ++k) {
 36         tree[t].min[k] = tree[t].max[k] = tree[t].pos[k];
 37         for (int i = 0, s; i < 2; ++i)if (s = tree[t].son[i]) {
 38             if (tree[t].min[k] > tree[s].min[k])
 39                 tree[t].min[k] = tree[s].min[k];
 40             if (tree[t].max[k] < tree[s].max[k])
 41                 tree[t].max[k] = tree[s].max[k];
 42         }
 43     }
 44 }
 45 
 46 int build(int l, int r, int k) {
 47     int d = (l + r) / 2; cmpk = k;
 48     nth_element(tree + l, tree + d, tree + r + 1);
 49     tree[d].son[0] = l == d ? 0 : build(l, d - 1, k ^ 1);
 50     tree[d].son[1] = r == d ? 0 : build(d + 1, r, k ^ 1);
 51     return update(d), d;
 52 }
 53 
 54 inline void build(void) {
 55     for (int i = 1; i <= n; ++i) {
 56         tree[i].id = i - 1;
 57         tree[i].pos[0] = next();
 58         tree[i].pos[1] = next();
 59     }
 60     root = build(1, n, 0);
 61 }
 62 
 63 int st[2], ed[2], ans[500005], tot = 0;
 64 
 65 inline bool judge(const node &t) {
 66     return t.pos[0] >= st[0]
 67         && t.pos[0] <= ed[0]
 68         && t.pos[1] >= st[1]
 69         && t.pos[1] <= ed[1];
 70 }
 71 
 72 inline bool check(const node &t) {
 73     return t.min[0] <= ed[0]
 74         && t.max[0] >= st[0]
 75         && t.min[1] <= ed[1]
 76         && t.max[1] >= st[1];
 77 }
 78 
 79 void query(int t) {
 80     if (judge(tree[t]))ans[tot++] = tree[t].id;
 81     if (tree[t].son[0] && check(tree[tree[t].son[0]]))query(tree[t].son[0]);
 82     if (tree[t].son[1] && check(tree[tree[t].son[1]]))query(tree[t].son[1]);
 83 }
 84 
 85 inline void query(void) {
 86     for (int i = 0; i < m; ++i, tot = 0) {
 87         st[0] = next();
 88         ed[0] = next();
 89         st[1] = next();
 90         ed[1] = next();
 91         tot = 0;
 92         query(root);
 93         sort(ans, ans + tot);
 94         for (int j = 0; j < tot; ++j)
 95             printf("%d\n", ans[j]);
 96         puts("");
 97     }
 98 }
 99 
100 signed main(void) {
101     n = next(); build();
102     m = next(); query();
103     //system("pause");
104 }

 

@Author: YouSiki

posted @ 2016-12-16 20:24  YouSiki  阅读(329)  评论(0)  编辑  收藏  举报