Same as LintCode "Sliding Window Median", but requires more care on details - no trailing zeroes.

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <cstdlib>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
using namespace std;
/* Head ends here */
multiset<int> lmax, rmin;
void removeOnly1(multiset<int> &ms, int v)
{
auto pr = ms.equal_range(v);
ms.erase(pr.first);
} void remove(multiset<int> &lmax, multiset<int> &rmin, int v)
{
if(v <= *lmax.rbegin())
{
removeOnly1(lmax, v);
if(lmax.size() < rmin.size())
{
int tmp = *rmin.begin();
lmax.insert(tmp);
removeOnly1(rmin, tmp);
}
}
else if(v >= *rmin.begin())
{
removeOnly1(rmin, v);
if((lmax.size() - rmin.size()) > )
{
int tmp = *lmax.rbegin();
removeOnly1(lmax, tmp);
rmin.insert(tmp);
}
}
} void addin(multiset<int> &lmax, multiset<int> &rmin, int v)
{
if(lmax.empty())
{
lmax.insert(v);
return;
}
int lmax_v = *lmax.rbegin();
int size_l = lmax.size(), size_r = rmin.size();
if(v <= lmax_v) // to add left
{
lmax.insert(v);
if((size_l + - size_r) > )
{
int tmp = *lmax.rbegin();
rmin.insert(tmp);
removeOnly1(lmax, tmp);
}
}
else
{
rmin.insert(v);
if((size_r + )> size_l)
{
int tmp = *rmin.begin();
removeOnly1(rmin, tmp);
lmax.insert(tmp);
}
}
} void median(vector<char> s,vector<int> X) {
int n = s.size();
multiset<int> ms;
for(int i = ; i < n; i ++)
{
if(s[i] == 'r')
{
if(!lmax.count(X[i]) && !rmin.count(X[i]))
{
cout << "Wrong!" << endl;
continue;
}
else
{
remove(lmax, rmin, X[i]);
}
}
else
{
addin(lmax, rmin, X[i]);
}
if(lmax.size() == rmin.size())
{
if(lmax.size() >)
{
long long f1 = (long long)(*lmax.rbegin());
long long f2 = (long long)(*rmin.begin());
if ((f1 + f2) % == ) {
printf("%.0lf\n",(f1*.+f2)/.);
}
else {
printf("%.1lf\n",(f1*.+f2)/.);
}
}
else
cout << "Wrong!" << endl;
}
else
{
printf("%d\n",*lmax.rbegin());
} } }
int main(void){ //Helpers for input and output int N;
cin >> N; vector<char> s;
vector<int> X;
char temp;
int tempint;
for(int i = ; i < N; i++){
cin >> temp >> tempint;
s.push_back(temp);
X.push_back(tempint);
} median(s,X);
return ;
}

HackerRank "Median Updates"的更多相关文章

  1. 【HackerRank】Median

    题目链接:Median 做了整整一天T_T 尝试了各种方法: 首先看了解答,可以用multiset,但是发现java不支持: 然后想起来用堆,这个基本思想其实很巧妙的,就是维护一个最大堆和最小堆,最大 ...

  2. 【HackerRank】Find the Median(Partition找到数组中位数)

    In the Quicksort challenges, you sorted an entire array. Sometimes, you just need specific informati ...

  3. Failure to find xxx in xxx was cached in the local repository, resolution will not be reattempted until the update interval of nexus has elapsed or updates are forced @ xxx

    问题: 在linux服务器上使用maven编译war时报错: 16:41:35 [FATAL] Non-resolvable parent POM for ***: Failure to find * ...

  4. No.004:Median of Two Sorted Arrays

    问题: There are two sorted arrays nums1 and nums2 of size m and n respectively.Find the median of the ...

  5. [LeetCode] Find Median from Data Stream 找出数据流的中位数

    Median is the middle value in an ordered integer list. If the size of the list is even, there is no ...

  6. [LeetCode] Median of Two Sorted Arrays 两个有序数组的中位数

    There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two ...

  7. Applying vector median filter on RGB image based on matlab

    前言: 最近想看看矢量中值滤波(Vector median filter, VMF)在GRB图像上的滤波效果,意外的是找了一大圈却发现网上没有现成的code,所以通过matab亲自实现了一个,需要学习 ...

  8. 【leetcode】Median of Two Sorted Arrays

    题目简述: There are two sorted arrays A and B of size m and n respectively. Find the median of the two s ...

  9. Codeforces Round #327 (Div. 2) B. Rebranding C. Median Smoothing

    B. Rebranding The name of one small but proud corporation consists of n lowercase English letters. T ...

随机推荐

  1. sass中mixin常用的CSS3

    圆角border-radius @mixin rounded($radius){ -webkit-border-radius: $radius; -moz-border-radius: $radius ...

  2. 获取验证码,60秒倒计时js

    <input type="button" id="btn" value="免费获取验证码" /><script type= ...

  3. poj3249 Test for Job ——拓扑+DP

    link:http://poj.org/problem?id=3249 在拓扑排序的过程中进行状态转移,dp[i]表示从起点到 i 这个点所得到的的最大值.比如从u点到v点,dp[v]=max(dp[ ...

  4. poj1062 最短路

    题意:有n个物品,任务是得到1号物品,现在每个物品有它的主人,你可以用金钱购买物品,当然也可以用其他物品加上优惠的价格换取,但是有个要求,因为每个物品的主人有各自的等级,你所交易过的人中,等级差不能超 ...

  5. 【NOIP2008】双栈排序

    感觉看了题解还是挺简单的,不知道当年chty同学为什么被卡了呢么久--所以说我还是看题解了 原题: Tom最近在研究一个有趣的排序问题.如图所示,通过2个栈S1和S2,Tom希望借助以下4种操作实现将 ...

  6. mave之:java的web项目必须要的三个jar的pom形式

    jsp-api javax.servlet-api jstl <!-- jsp --> <dependency> <groupId>javax.servlet< ...

  7. 论文笔记之: Bilinear CNN Models for Fine-grained Visual Recognition

    Bilinear CNN Models for Fine-grained Visual Recognition CVPR 2015 本文提出了一种双线性模型( bilinear models),一种识 ...

  8. JSBinding + SharpKit / 常见问题

    运行时出现: Return a "System.Xml.XmlIteratorNodeList" to JS failed. Did you forget to export th ...

  9. 关于CPU Cache -- 程序猿需要知道的那些事

    本文将介绍一些作为程序猿或者IT从业者应该知道的CPU Cache相关的知识 文章欢迎转载,但转载时请保留本段文字,并置于文章的顶部 作者:卢钧轶(cenalulu) 本文原文地址:http://ce ...

  10. 动态加载dll,并创建类对象放入到list中。

    private List<IVisualControlsPlug> visualPlugs = new List<IVisualControlsPlug>(); public ...