线段树+离散化 IP地址段检查 SEGMENT TREE

最新推荐文章于 2023-12-23 10:32:54 发布

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本文介绍了一种通过构建区间树来高效查找IP地址所属段的方法。首先将IP段离散化为整数区间，然后建立区间树，每个叶节点对应一个具体的IP区间。通过递归插入和搜索操作，实现快速定位特定IP地址所在的段。

Problem:

Give a series of IP segments, for example, [0.0.0.1-0.0.0.3], [123.234.232.21-123.245.21.1]...

Now there is a new IP, find which IP segment it's in ?

Solution:

First, we could map the ends of IP segments into some intervals, since the IP address could be represented by a unsigned int. This is called discretization.

Second, setup the segment tree. Every leaf node is [x, x+1], it means whether an IP segment includes the part. We could use a vector<int> to record the interval index for searching. So the core code is:

#include <iostream>
#include <map>
#include <vector>
#include <algorithm>
#include <limits.h>
#include <assert.h>
#include <stdio.h>
using namespace std;

class Interval {
 public:
  int l, r;
  Interval(int ll = 0, int rr = 0) : l(ll), r(rr) {}
};
class Node {
 public:
  int l, r;
  Node *lc, *rc;
  vector<int> IPs;
  
  Node(int ll = 0, int rr = 0) : l(ll), r(rr), lc(NULL), rc(NULL) {}
};

Node* build(int l, int r) {
  Node *root = new Node();
  root->l = l, root->r = r;
  if (l + 1 >= r)
    return root;
  int mid = (l + r) / 2;
  root->lc = build(l, mid);
  root->rc = build(mid, r);
  return root;
}
void insert(Node* root, int l, int r, int idx) {
  if (l <= root->l && r >= root->r) {
    root->IPs.push_back(idx);
    return;
  }
  int mid = (root->l + root->r) / 2;
  if (l < mid) // Take care here!
    insert(root->lc, l, mid, idx);
  if (r > mid) // Take care here!
    insert(root->rc, mid, r, idx);
}
void search(Node* root, int l, int r, vector<int>& res) {
  if (r <= root->r && l >= root->l) {
    for (int i = 0; i < root->IPs.size(); ++i)
      res.push_back(root->IPs[i]);
  }
  int mid = (root->l + root->r) / 2;
  if (r <= mid)
    search(root->lc, l, mid, res);
  if (l >= mid)
    search(root->rc, mid, r, res);
}
int main()
{
  Node* rt = build(1, 7);
  Interval inters[] = {Interval(1,3),Interval(2,4),Interval(1,5), Interval(5,7)};
  int size = sizeof(inters) / sizeof(inters[0]);
  for (int i = 0; i < size; ++i) 
    insert(rt, inters[i].l, inters[i].r, i);
  vector<int> res;
  search(rt, 2, 3, res);
  return 0;
}