Submission #84189

#	TimeUTC-0	Username	Problem	Language	Result	Execution time	Memory
84189	2018-11-13 21:05:07	radoslav11	New Home (APIO18_new_home)	C++14	100 / 100	4417 ms	837060 KiB

new_home

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/*
   We will use sweep line to solve the problem. We split the stores into 2 queries: 
   1) Add store i at time a[i]
   2) Remove store i at time b[i] + 1
   We will also have queries in the sweep line. Everything will be sorted by time in increasing order.
   Now to handle queries we will maintain K sets - the available positions of j-type stores. Then if A and B are two consecutive stores, the closest elements to all positions in [A; B] are A or B.
   Then let's have a two segment trees wtih sets - one for closest elements to the left and one for closest elements to the right. Now addition of store with type X will be done like that:
   1) Let A <= X <= B and A and B are the closest stores of the same type. 
   2) We remove the interval [A; B] from the DS.
   3) We add the intervals [A; X] and [X; B].
   Adding or removing an interval is done by finding the middle position and then concidering the two ranges - [L; Mid] and [Mid + 1; R].
   The complexity will be O(N * log N * log N).
   As sets are slow, we will compress the input in each segment tree node beforehand and then use priority queue instead of sets.
   Unfortunately the above data structure was too slow. So my second idea is to change the data structure to two simple treaps and do binary search on them. 
   The complexity will be O(N log N) this way. 
   Again unfortunately the treap solution was too slow (it got 47). So the third idea is to make the data structure offline. Then the treap can be replaced with segment tree.
   */
#include <bits/stdc++.h>
#define endl '\n'
//#pragma GCC optimize ("O3")
//#pragma GCC target ("sse4")
 
 
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Compilation message (stderr)

new_home.cpp: In function 'void next_char()':
new_home.cpp:513:70: warning: ignoring return value of 'size_t fread(void*, size_t, size_t, FILE*)', declared with attribute warn_unused_result [-Wunused-result]
 void next_char() { if(++pos_buff == maxl) fread(buff, 1, maxl, stdin), pos_buff = 0; }
                                           ~~~~~~~~~~~~~~~~~~~~~~~~~~~^~~~~~~~~~~~~~

• Subtask 1: 5 / 5
• Subtask 2: 7 / 7
• Subtask 3: 10 / 10
• Subtask 4: 23 / 23
• Subtask 5: 35 / 35
• Subtask 6: 20 / 20