Segment tree + discrete IP address SEGMENT check SEGMENT TREE

Source: Internet
Author: User

Segment tree + discrete IP address SEGMENT check SEGMENT TREE


Give a series of IP segments, for example, [], []...

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


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

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

    #include #include 
     #include #include 
      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
        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
        & 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
          res; search(rt, 2, 3, res); return 0;}

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