线段树+离散化 IP地址段检查 SEGMENT TREE

来源：互联网发布：矮人村庄mac 编辑：程序博客网时间：2024/05/13 15:15

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;}

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