AGC014E Blue and Red Tree

题意

There is a tree with (N) vertices numbered (1) through (N). The (i)-th of the (N−1) edges connects vertices (a_i) and (b_i).

Initially, each edge is painted blue. Takahashi will convert this blue tree into a red tree, by performing the following operation (N−1) times:

• Select a simple path that consists of only blue edges, and remove one of those edges.
• Then, span a new red edge between the two endpoints of the selected path.

His objective is to obtain a tree that has a red edge connecting vertices (c_i) and (d_i), for each (i).

Determine whether this is achievable.

分析

这题可以倒着想，在原图中最后删掉的边，在新图中也一定是以同样的方式出现。

那么我们可以看成一开始原图中有n个点连通块，每次找一条在原图和新图中都出现的边，把这条边的两个端点缩成一个点，然后继续进行以上操作。

若还未合并至一点且无法继续合并则不可行。

用并查集维护连通块，map维护边出现的次数，set维护点的邻接点，用queue维护当前可合并的点对。合并时需要找到第一个不在同一连通块中的可合并点对（因为以前的合并可能让它们属于同一连通块），find一下合并连通块的并查集的根，然后用启发式合并把x的邻接点赋给y，删掉x的一切记录（省空间），最后把出现次数改变成2的边放入queue中。

代码

#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<ctime>
#include<iostream>
#include<string>
#include<vector>
#include<list>
#include<deque>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<algorithm>
#include<complex>
#pragma GCC optimize ("O0")
using namespace std;
template<class T> inline T read(T&x){
    T data=0;
	int w=1;
    char ch=getchar();
    while(!isdigit(ch))
    {
		if(ch=='-')
			w=-1;
		ch=getchar();
	}
    while(isdigit(ch))
        data=10*data+ch-'0',ch=getchar();
    return x=data*w;
}
typedef long long ll;
typedef pair<int,int>pii;

const int MAXN=1e5+7;

int n;
map<ll,int>M;
queue<pii>Q;
set<int>l[MAXN];
int fa[MAXN];

int find(int x)
{
	return fa[x]==x?x:fa[x]=find(fa[x]);
}

ll getid(int x,int y)
{
	if(x>y)
		swap(x,y);
	return (ll)x*n+y;
}

void link(int x,int y)
{
	l[x].insert(y);
	l[y].insert(x);
	ll v=getid(x,y);
	++M[v];
	if(M[v]==2)
		Q.push(pii(x,y));
}

int main()
{
//  freopen(".in","r",stdin);
//  freopen(".out","w",stdout);
	read(n);
	for(int i=1;i<=n;++i)
		fa[i]=i;
	for(int i=1;i<=2*n-2;++i)
	{
		int x,y;
		read(x);read(y);
		link(x,y);
	}
	for(int i=1;i<n;++i)
	{
		int x,y;
		while(1)
		{
			if(Q.empty())
			{
				printf("NO
");
				return 0;
			}
			x=Q.front().first,y=Q.front().second;
			Q.pop();
			x=find(x),y=find(y);
			if(x!=y)
				break;
		}
		if(l[x].size()>l[y].size())
			swap(x,y);
		fa[x]=y;
		M.erase(getid(x,y));
		l[y].erase(x);
		for(auto v:l[x])
		{
			v=find(v);
			if(v==y)
				continue;
			M.erase(getid(x,v));
			l[v].erase(x);
			l[x].erase(v);
			link(v,y);
		}
	}
	printf("YES
");
//  fclose(stdin);
//  fclose(stdout);
    return 0;
}