线段树+离散化 IP地址段检查 SEGMENT TREE
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;
}
线段树+离散化 IP地址段检查 SEGMENT TREE的更多相关文章
- POJ2528Mayor's posters[线段树 离散化]
Mayor's posters Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 59683 Accepted: 17296 ...
- POJ1151Atlantis 矩形面积并[线段树 离散化 扫描线]
Atlantis Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 21734 Accepted: 8179 Descrip ...
- POJ 2528 Mayor's posters(线段树+离散化)
Mayor's posters 转载自:http://blog.csdn.net/winddreams/article/details/38443761 [题目链接]Mayor's posters [ ...
- [poj2528] Mayor's posters (线段树+离散化)
线段树 + 离散化 Description The citizens of Bytetown, AB, could not stand that the candidates in the mayor ...
- BZOJ_4653_[Noi2016]区间_线段树+离散化+双指针
BZOJ_4653_[Noi2016]区间_线段树+离散化+双指针 Description 在数轴上有 n个闭区间 [l1,r1],[l2,r2],...,[ln,rn].现在要从中选出 m 个区间, ...
- D - Mayor's posters(线段树+离散化)
题目: The citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campai ...
- POJ 1151Atlantis 矩形面积并[线段树 离散化 扫描线]
Atlantis Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 21734 Accepted: 8179 Descrip ...
- hdu1542 矩形面积并(线段树+离散化+扫描线)
题意: 给你n个矩形,输入每个矩形的左上角坐标和右下角坐标. 然后求矩形的总面积.(矩形可能相交). 题解: 前言: 先说说做这道题的感受: 刚看到这道题顿时就懵逼了,几何 烂的渣渣.后来从网上搜题解 ...
- Mayor's posters (线段树+离散化)
Mayor's posters Description The citizens of Bytetown, AB, could not stand that the candidates in the ...
随机推荐
- Android中 string.xml资源 如何添加参数?
在android 开发,我们通常会用string.xml资源去设置textview等控件的字符串.而值一般是与程序的运行结果无关的. 但有时需要根据运行的结果来显示到控件中,这时字符串资源就不能写死了 ...
- 使用Spring框架的步骤
“好记性,不如烂笔头”.今天正式接触了Spring框架,第一次接触Spring框架感觉Spring框架简化了好多程序代码,开发效率大大提高.现在介绍使用Spring框架的步骤.(使用spring-fr ...
- HDU_2079_(01背包)(dfs)
选课时间(题目已修改,注意读题) Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) ...
- CAD使用SetXData写数据(网页版)
主要用到函数说明: MxDrawEntity::SetXData 设置实体的扩展数据,详细说明如下: 参数 说明 [in] IMxDrawResbuf* pXData 扩展数据链表 js代码实现如下: ...
- Java基础概念语法
Java基础概念语法 注释 单行注释 //行注释说明 多行注释 /* 多行注释说明 */ 文档注释 /** *@author 程序的作者 *@version 源文件的版本 *@param 方法的参数说 ...
- 15Microsoft SQL Server 数据库维护
Microsoft SQL Server 数据库维护 2.6.1数据库联机与脱机 --联机:该状态为数据库正常状态,也就是我们常看到的数据库的状态,该状态下的数据库处于可操作状态,可以对数据库进行任何 ...
- MongoDB 启动和关闭
重启命令 service mongodb restart 启动命令 mongod -f /data/tools/mongodb/config/config.conf 必须要带配置文件才能启动 关闭命令 ...
- 【loj6184】无心行挽(虚树+倍增)
题目链接:https://loj.ac/problem/6184 每次询问给一些关键点,询问树上每个点离最近的关键点的距离(以后称为f(u))最大值是多少. 询问数比较大,但 \sum{K} 和n是一 ...
- POJ P2096 Collecting Bugs
思路 分类讨论,不妨先设$DP[i][j]$表示已经发现$i$种子系统中有$n$种$bug$无非只有四种情况 发现的$bug$在旧的系统旧的分类,概率$p1$是$(i/s)*(j/n)$. 发现的$b ...
- 爬虫之pyquery库
官方文档:https://pyquery.readthedocs.io/en/latest/ PyQuery是一个强大又灵活的网页解析库.如果你觉得正则写起来太麻烦.BeautifulSoup语法太难 ...