제출 #880338

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
880338		Lib	Split the Attractions (IOI19_split)	C++14	0 / 100	1 ms	2396 KiB

split

#include <bits/stdc++.h>
#include "split.h"
using namespace std;
vector <vector <int> > adj;
vector <int> TVector;
int cpos[200003];
int check[200003];
int SubtreeSize[200003];
int Parent[200003];
int n;
//Parent on each iteration of DFS
deque <int> clist;
vector <int> ans;
	vector <int> Group1,Group2;
	int GroupAID,GroupBID,GroupASize,GroupBSize;
	int cur;
int Subtree1,Subtree2;
void ShittyDFS(int i){
	for(int k=0;k<n;k++){
			cpos[k]=0;
			SubtreeSize[k]=0;
			Parent[k]=0;
		}
		Parent[i]=-1;
		//wacky dfs (and calculate size of subtrees as well
		cur=1;
		clist.push_back(i);
		check[i]=1;
		while(!clist.empty()){
			if(!SubtreeSize[clist.back()]){
				SubtreeSize[clist.back()]=cur;
			}
			while(cpos[clist.back()]<adj[clist.back()].size()&&check[adj[clist.back()][cpos[clist.back()]]]){
				cpos[clist.back()]++;
			}
			if(cpos[clist.back()]<adj[clist.back()].size()&&!check[adj[clist.back()][cpos[clist.back()]]]){
				check[adj[clist.back()][cpos[clist.back()]]]=1;
				Parent[adj[clist.back()][cpos[clist.back()]]]=clist.back();
				clist.push_back(adj[clist.back()][cpos[clist.back()]]);
				cur++;
			}else{
				SubtreeSize[clist.back()]=cur-SubtreeSize[clist.back()];
				clist.pop_back();
			}
		}
}
vector <int> find_split(int n, int a, int b, int c, vector <int> p, vector <int> q){
	//Only cares about the 2 smaller group: Assuming that we DID find a solution where the largest group is connected, just swap some of those points for the unused group.
	//As this is the largest group, swapping some minor connected parts for the unused group is still valid, and always doable => only care about 2 smallest group only
	//very intuitive tbh
	vector <pair <int,int> > order;
	order.push_back({a,1});
	order.push_back({b,2});
	order.push_back({c,3});
	sort(order.begin(),order.end());
	GroupAID=order[0].second;
	GroupASize=order[0].first;
	GroupBID=order[1].second;
	GroupBSize=order[1].first;
	for(int i=0;i<=n+1;i++){
		adj.push_back(TVector);
	}
	//Build the graph. Who can't understand this?
	for(int i=0;i<p.size();i++){
		adj[p[i]].push_back(q[i]);
		adj[q[i]].push_back(p[i]);
	}
	/*
	Let's just call the smallest group gropu A, and second smallest group group B.
	Let's kinda start working with the tree subtask first. When we split the tree into 2 part, we are essentially dedicating an entire subtree for group A and the rest of the
	tree for group B, or vice versa. We will now do the following:\
	- Try all vertices on the tree. For each verticies, check through all of its subtreees.
	- If we can dedicate an entire subtree to group A and the rest to group B (or vice versa), this split will work instantly. We are basically cutting our original tree into 2 parts.
	- If after checking all subtrees of all vertices and we still haven't found a solution, there is no solution.
	(Why this kind of check works: We are trying everything, so no method of "splitting OG tree into 2 parts" is ignored...Let's be honest, I think that 64pts for this is trivial and
	doesn't require much thought at all. The wacky funny parts comes from the final subtask, which I haven't solved yet.
	Complexity: We take O(N) to DFS and calculate size of subtrees of all nodes. For each of the N vertex, we check through all of their adjacient vertices. Basically, we are checking
	3N times (like, come on, in a tree of N nodes, there are 2N-2 pairs of adjacient points. There is literally nothing to explain here). Basically, for each DFS tree, we take roughly
	O(N) to check whether there exists a valid split or not.
	For the general graph (but small) case: Just create N DFS trees and bruteforce them, lol. Nothing to talk about just yet
	*/
	for(int i=0;i<n;i++){
		//Tested locally, seems correct enough. First time implementing DFS. Moving on.
		//On another hand, you know what? Chugging DFS into a function now. brb
		
		//Done
		ShittyDFS(i);
		//Try the split with every single vertex on the tree.
		for(int k=0;k<n;k++){
			for(int l=0;l<adj[k].size();l++){
				Subtree1=SubtreeSize[adj[k][l]];
				if(adj[k][l]==Parent[k]){
					Subtree1=n-SubtreeSize[k];
				}
				Subtree2=n-Subtree1;
				if(Subtree1>=GroupASize&&Subtree2>=GroupBSize){
					check[adj[k][l]]=GroupAID;
					check[k]=GroupBID;
					clist.push_back(adj[k][l]);
					while(clist.size()<GroupASize){
						
					}
				}
			}
		}
	}
	return ans;
}

컴파일 시 표준 에러 (stderr) 메시지

split.cpp: In function 'void ShittyDFS(int)':
split.cpp:33:28: warning: comparison of integer expressions of different signedness: 'int' and 'std::vector<int>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
   33 |    while(cpos[clist.back()]<adj[clist.back()].size()&&check[adj[clist.back()][cpos[clist.back()]]]){
      |          ~~~~~~~~~~~~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~
split.cpp:36:25: warning: comparison of integer expressions of different signedness: 'int' and 'std::vector<int>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
   36 |    if(cpos[clist.back()]<adj[clist.back()].size()&&!check[adj[clist.back()][cpos[clist.back()]]]){
      |       ~~~~~~~~~~~~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~
split.cpp: In function 'std::vector<int> find_split(int, int, int, int, std::vector<int>, std::vector<int>)':
split.cpp:64:15: warning: comparison of integer expressions of different signedness: 'int' and 'std::vector<int>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
   64 |  for(int i=0;i<p.size();i++){
      |              ~^~~~~~~~~
split.cpp:90:17: warning: comparison of integer expressions of different signedness: 'int' and 'std::vector<int>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
   90 |    for(int l=0;l<adj[k].size();l++){
      |                ~^~~~~~~~~~~~~~
split.cpp:100:24: warning: comparison of integer expressions of different signedness: 'std::deque<int>::size_type' {aka 'long unsigned int'} and 'int' [-Wsign-compare]
  100 |      while(clist.size()<GroupASize){
      |            ~~~~~~~~~~~~^~~~~~~~~~~

• Subtask 1: 0 / 7
• Subtask 2: 0 / 11
• Subtask 3: 0 / 22
• Subtask 4: 0 / 24
• Subtask 5: 0 / 36