CodeForces - 1559D2 Mocha and Diana (Hard Version)(思维)

Posted 2021-09-07 Frozen_Guardian

题目链接：点击查看

题目大意：给出两棵森林，每次可以同时在两个森林中增加同一条边，问最多可以增加多少条边，使得两个森林仍然还是森林

题目分析：结论参考至：https://blog.csdn.net/RunningBeef/article/details/119738571?utm_source=app&app_version=4.13.0&code=app_1562916241&uLinkId=usr1mkqgl919blen

首先对于所有的点尝试与点 $1$ 建边，此后考虑除了点 $1$ 以外的联通块。假设两个森林中的联通块集合分别为 $s1$ 和 $s2$，那么令 $s1$ 中的连通块和 $s2$ 中的任意一个连通块相连都是合法的

代码：

// Problem: D2. Mocha and Diana (Hard Version)
// Contest: Codeforces - Codeforces Round #738 (Div. 2)
// URL: https://codeforces.com/contest/1559/problem/D2
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast","inline","-ffast-math")
// #pragma GCC target("avx,sse2,sse3,sse4,mmx")
#include<iostream>
#include<cstdio>
#include<string>
#include<ctime>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<climits>
#include<queue>
#include<map>
#include<set>
#include<sstream>
#include<cassert>
#include<bitset>
#include<list>
#include<unordered_map>
#define lowbit(x) (x&-x)
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
template<typename T>
inline void read(T &x)
{
	T f=1;x=0;
	char ch=getchar();
	while(0==isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
	while(0!=isdigit(ch)) x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
	x*=f;
}
template<typename T>
inline void write(T x)
{
	if(x<0){x=~(x-1);putchar('-');}
    if(x>9)write(x/10);
    putchar(x%10+'0');
}
const int inf=0x3f3f3f3f;
const int N=1e6+100;
int f1[N],f2[N];
int find1(int x) {
	return f1[x]==x?x:f1[x]=find1(f1[x]);
}
int find2(int x) {
	return f2[x]==x?x:f2[x]=find2(f2[x]);
}
bool check1(int x,int y) {
	int xx=find1(x),yy=find1(y);
	if(xx!=yy) {
		return true;
	}
	return false;
}
bool check2(int x,int y) {
	int xx=find2(x),yy=find2(y);
	if(xx!=yy) {
		return true;
	}
	return false;
}
void merge1(int x,int y) {
	int xx=find1(x),yy=find1(y);
	f1[xx]=yy;
}
void merge2(int x,int y) {
	int xx=find2(x),yy=find2(y);
	f2[xx]=yy;
}
void init() {
	for(int i=0;i<N;i++) {
		f1[i]=f2[i]=i;
	}
}
int main()
{
#ifndef ONLINE_JUDGE
//	freopen("data.in.txt","r",stdin);
//	freopen("data.out.txt","w",stdout);
#endif
//	ios::sync_with_stdio(false);
	init();
	int n,m1,m2;
	cin>>n>>m1>>m2;
	for(int i=1;i<=m1;i++) {
		int u,v;
		read(u),read(v);
		merge1(u,v);
	}
	for(int i=1;i<=m2;i++) {
		int u,v;
		read(u),read(v);
		merge2(u,v);
	}
	vector<pair<int,int>>ans;
	vector<int>rt1,rt2;
	for(int i=2;i<=n;i++) {
		if(check1(1,i)&&check2(1,i)) {
			merge1(1,i),merge2(1,i);
			ans.push_back({1,i});
		}
	}
	for(int i=2;i<=n;i++) {
		if(find1(i)==i&&check1(1,i)) {
			rt1.push_back(i);
		}
		if(find2(i)==i&&check2(1,i)) {
			rt2.push_back(i);
		}
	}
	for(int i=0;i<(int)min(rt1.size(),rt2.size());i++) {
		ans.push_back({rt1[i],rt2[i]});
	}
	cout<<ans.size()<<endl;
	for(auto it:ans) {
		cout<<it.first<<' '<<it.second<<endl;
	}
	return 0;
}