제출 #235222

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
235222		rajarshi_basu	조이터에서 친구를 만드는건 재밌어 (JOI20_joitter2)	C++14	0 / 100	13 ms	15616 KiB

joitter2

#include <stdio.h>     
#include <stdlib.h>    
#include <iostream>
#include <vector>
#include <algorithm>
#include <fstream>
#include <queue>
#include <deque>
#include <iomanip>
#include <cmath>
#include <set>
#include <stack>
#include <map>
#include <unordered_map>

#define FOR(i,n) for(int i=0;i<n;i++)
#define FORE(i,a,b) for(int i=a;i<=b;i++)
#define ll long long 
#define ld long double
//#define int short
#define vi vector<int>
#define pb push_back
#define ff first
#define ss second
#define ii pair<int,int>
#define iii pair<int,ii>
#define pll pair<ll,ll>
#define plll pair<ll,pll>
#define mp make_pair
#define vv vector
#define endl '\n'

using namespace std;

/*
1) when merging two cliques of size A and B, we need to add A*B-1 edges.
*/


const int MAXN = 1e5+5;
int D = 0;
int E = 0;
int F = 0;
int G = 0;
struct DSU{
	ll tot = 0;
	int parent[MAXN];
	set<int> incoming[MAXN];// individual nodes 
	set<int> incomingCliques[MAXN];
	set<int> allIncomers;
	set<ii> outgoing[MAXN];//
	ll sz[MAXN];
	DSU(){
		FOR(i,MAXN)parent[i] = i;
		//FOR(i,MAXN)incoming[i] = new set<int>(), outgoing[i] = new set<ii>();
		FOR(i,MAXN)sz[i] = 1;
	}
	int find(int a){
		if(a != parent[a])parent[a] = find(parent[a]);
		return parent[a];
	}
	bool isSame(int a,int b){
		return find(a) == find(b);
	}
	// outgoing[e] = {a,i} means there are i edges going from Supernodes e to a
	// we only keep track of outgoing edges + internal clique edges;


	inline void addEdge(int a,int b){
		b = find(b);
		incomingCliques[b].insert(find(a));
		incoming[b].insert(a);
		tot += sz[b];
	}

	void makeParent(int pa,int pb){
		// make pa parent
		allIncomers.clear();
		if(F)cout << "tot : " << tot << endl;
		if(G){
			cout << pa << " <> " << pb << endl;
			for(auto e: incoming[pa])cout << e << " ";cout << endl;
			for(auto e: incoming[pb])cout << e << " ";cout << endl;
		}
		for(auto e : incoming[pa]){
			if(find(e) != pb)
				allIncomers.insert(e);
			tot -= sz[pa];
		}

		for(auto e : incoming[pb]){
			if(find(e) != pa)
				allIncomers.insert(e);
			tot -=sz[pb];
		}
		incoming[pa].clear();
		incoming[pb].clear();
		incomingCliques[pa].clear();
		incomingCliques[pb].clear();
		// destroyed all incoming edges. Now, we will add them back. 
		
		if(F)cout << "tot : " << tot << endl;
		
	}


	void merge(int a,int b){
		if(isSame(a,b))return;
		int pa = find(a);
		int pb = find(b);
		tot += 2*sz[pa]*sz[pb]; // new internal edges of the clique.

		if(sz[a] > sz[b]){
			makeParent(pa,pb);
			parent[pb] = pa;
			sz[pa] += sz[pb];
		}else{
			makeParent(pb,pa);
			parent[pa] = pb;
			sz[pb] += sz[pa];
		}
		for(auto e : allIncomers){
			//cout << sz[b];
			//cout << "EDGE ADDED EXPLICITLY: " << e << endl;
			addEdge(e,pa);
		}
	}

	void handleEdge(int a,int b){
		int pa = find(a);
		int pb = find(b);
		// edge is a-> b
		if(incoming[pb].find(a) != incoming[pb].end())return;
		if(incomingCliques[pa].find(pb) != incomingCliques[pa].end()){
			//need to merge cliques
			if(E)cout << "NEED TO MERGE" << endl;
			merge(a,b);
		}else{
			addEdge(a,b);
		}
	}
};	

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);

	int n,m;
	cin >> n >> m;
	DSU dsu;
	FOR(i,m){
		int a,b;
		cin >> a >> b;
		a--;b--;
		dsu.handleEdge(a,b);
		cout << dsu.tot << endl;
	}
	return 0;
}

컴파일 시 표준 에러 (stderr) 메시지

joitter2.cpp: In member function 'void DSU::makeParent(int, int)':
joitter2.cpp:82:4: warning: this 'for' clause does not guard... [-Wmisleading-indentation]
    for(auto e: incoming[pa])cout << e << " ";cout << endl;
    ^~~
joitter2.cpp:82:46: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the 'for'
    for(auto e: incoming[pa])cout << e << " ";cout << endl;
                                              ^~~~
joitter2.cpp:83:4: warning: this 'for' clause does not guard... [-Wmisleading-indentation]
    for(auto e: incoming[pb])cout << e << " ";cout << endl;
    ^~~
joitter2.cpp:83:46: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the 'for'
    for(auto e: incoming[pb])cout << e << " ";cout << endl;
                                              ^~~~

• Subtask 1: 0 / 1
• Subtask 2: 0 / 16
• Subtask 3: 0 / 83