제출 #874541

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
874541		Lib	Roller Coaster Railroad (IOI16_railroad)	C++14	30 / 100	74 ms	10928 KiB

railroad

#include <bits/stdc++.h>
#include "railroad.h"
using namespace std;
int n;
vector <pair <int,int> > EnterSpeed;
vector <pair <int,int> > ExitSpeed;
long long plan_roller_coaster (vector <int> s, vector <int> t){
	n=s.size();
	EnterSpeed.clear();
	ExitSpeed.clear();
	/*
	Subtask 1 and 2 is trival and basically not worth spending time for, so I'll skip it
	Of course, there exists a better solution to subtask 3 that is actually relevant to the full solution, but 30 points is 30 points,
	especially if that's from a shitty greedy solution
	*/
	
	/*
	If a finished track could be created without adding in any additional segments, this is the equivalent of: for all but 1 segment, there
	has to be a track segment that follows it.
	=> Just greedily assign exit speeds with corresponding entrance speeds that isn't from the same segment,
	if n-1 pairs (or more) could be formed, there exists an answer.
	Simple as that.
	*/
	for(int i=0;i<n;i++){
		EnterSpeed.push_back({s[i],i});
		ExitSpeed.push_back({t[i],i});
	}
	sort(EnterSpeed.begin(),EnterSpeed.end(),greater<pair <int,int> >());
	sort(ExitSpeed.begin(),ExitSpeed.end(),greater<pair <int,int> >());
	int ValidPairs=0;
	pair <int,int> temp;
	//Assigning track endpoints with new tracks
	while(!EnterSpeed.empty()&&!ExitSpeed.empty()&&ValidPairs<n-1){
		//Trying to match with the startpoint that isn't in the same track, and has the smallest entrance speed possible
		while(!EnterSpeed.empty()&&ExitSpeed.back().first>EnterSpeed.back().first){
			EnterSpeed.pop_back();
		}
		//It matches and doesn't belong to the same sector. Is valid, so increase the amount of matched pairs by 1
		if(!EnterSpeed.empty()&&ExitSpeed.back().second!=EnterSpeed.back().second){
			ValidPairs++;
			EnterSpeed.pop_back();
			ExitSpeed.pop_back();
		}else if(EnterSpeed.size()>1){
			//It matches, but both of this startpoint and endpoint belongs in the same segment. Match it with the
			//next smallest unmatched entrance speed
			temp=EnterSpeed.back();
			EnterSpeed.pop_back();
         	EnterSpeed.pop_back();
			ExitSpeed.pop_back();
			ValidPairs++;
			EnterSpeed.push_back(temp);
		}else{
          //If you can't find any other segment and ran into a conflict, all is lost and done. Stop the assigning 
          //process immediately and hope for the best
          break;
        }
	}
	//If at least N-1 pairs could be matched, there exists 1 valid rollercoaster using only the default tracks. Otherwise, no
	return (ValidPairs<n-1);
}
/*
Further update: this greedy algorithm is pretty much outright wrong.
We could think like this: Whenever we connect the end of track i with the beginning of track j, we are essentially creating an edge
from vertex i to vertex j, if we're thinking abouut the set of tracks as vertices on a graph.
The resulting "graph" of a valid track allocation should be either a "chain"/line/whatever, or a singular cycle. This kind of greedy 
assignment wouldn't always result in a valid graph.
THe created graph is only guaranteed to be a graph with n vertices, n/n-1 edges, and each vertex only have an in and out degree of 0 or 1.
Basically, it could create a graph with several cycles and a chain, instead of a singular chain or a singular cycle. Take this test as
an example:
5
2 11
8 3
4 7
10 5
6 9
(stolen from Codeforces).
Ideally, the track allocation should be a permutation of (1,2,3,4,5). However, whatever tf you do, this greedy algorithm would result in
2 cycles (2=>3,3=>2), (4=>5,5=>4) and the 1st track is left alone. Turns out the test for this problem is shitty, lmao. Imagine passing
with a wrong greedy algorithm. They should've seen it coming tbh.
.......30 free pts in something this hard is still funny doe. This *is* a shitty greedy, after all
*/

• Subtask 1: 0 / 11
• Subtask 2: 0 / 23
• Subtask 3: 30 / 30
• Subtask 4: 0 / 36