[ICPC2015WF] Tours

题目描述

The Arca Carania Mountain national park is opening up for tourist traffic. The national park has a number of sites worth seeing and roads that connect pairs of sites. The park commissioners have put together a set of round tours in the park in which visitors can ride buses to view various sites. Each round tour starts at some site (potentially different sites for different tours), visits a number of other sites without repeating any, and then returns to where it started. At least 3 different sites are visited in each round tour. At least one round tour is possible in the national park.

The park commissioners have decided that, for any given road, all buses will be operated by a single company. The commissioners do not want to be accused of favoritism, so they want to be sure that each possible round tour in the park has exactly the same number of roads assigned to each bus company. They realize this may be difficult to achieve. Thus, they want to learn what numbers of bus companies allow for a valid assignment of companies to roads.

Consider Sample Input 1, which is illustrated in Figure 1. There are a total of three round tours for these sites. Some company is assigned road 1-3. It must also be assigned some road on the round tour 1-2-3-4-1, say 2-3. But then it is assigned to two of the three roads on the round tour 1-2-3-1, and no other company can match this – so there can be no other companies. In Sample Input 2 there is only one round tour, so it is enough to assign the roads of this tour equally between companies.

Figure 1: Sample Input 1.

输入格式

The first line of input contains two integers \(n\) (\(1 \le n \le 2\, 000\)), which is the number of sites in the park, and \(m\) (\(1 \le m \le 2\, 000\)), which is the number of roads between the sites. Following that are \(m\) lines, each containing two integers \(a_ i\) and \(b_ i\) (\(1 \leq a_ i < b_ i \leq n\)), meaning the sites \(a_ i\) and \(b_ i\) are connected by a bidirectional road. No pair of sites is listed twice.

输出格式

Display all integers \(k\) such that it is possible to assign the roads to \(k\) companies in the desired way. These integers should be in ascending order.

盲猜答案是某个数的所有因数。

如果现在有两个环长度分别为 \(x,y\)，他们相交了 \(d\) 个点，那么 \(k|x,y,d\)

扩展一下，设 \(C(S)\) 为被且只被 \(S\) 中的环覆盖的边有 \(C(S)\) 条，那么 \(k|C(S)\)

我们现在就是要求出他们的 gcd.

然后跑出原图的一个生成树，对于一棵不在树上的边 \((u,v)\) ，那么在树上的 \((u,v)\) 这条路径可以全部分裂出来。可以用哈希维护，然后权值不同的边就是属于不同的等价类。同时给不在树上的边随机一个权值 \(w\)，并给 \((u,v)\) 这条路径全部异或上 \(w\)。

#include<bits/stdc++.h>
using namespace std;
const int N=2005;
int hd[N],n,m,vs[N],w[N],g[N],u[N],v[N],fa[N][24],f[N],dep[N],d,e_num;
map<int,int>c;
mt19937 gen(time(0));
struct edge{
	int v,nxt;
}e[N<<1];
int gcd(int x,int y)
{
	if(!y)
		return x;
	return gcd(y,x%y);
}
void add_edge(int u,int v)
{
	e[++e_num]=(edge){v,hd[u]};
	hd[u]=e_num;
	e[++e_num]=(edge){u,hd[v]};
	hd[v]=e_num;
}
void dfs(int x,int y)
{
	fa[x][0]=y;
	vs[x]=1;
	dep[x]=dep[y]+1;
	for(int i=1;i<=20;i++)
		fa[x][i]=fa[fa[x][i-1]][i-1];
	for(int i=hd[x];i;i=e[i].nxt)
		if(e[i].v^y)
			dfs(e[i].v,x);
}
void sou(int x,int y)
{
	vs[x]=0;
	for(int i=hd[x];i;i=e[i].nxt)
		if(e[i].v^y)
			sou(e[i].v,x),g[x]^=g[e[i].v];
	if(g[x])
		c[g[x]]++;
}
int find(int x)
{
	if(f[x]==x)
		return x;
	return f[x]=find(f[x]);
}
int lca(int x,int y)
{
	if(dep[x]<dep[y])
		swap(x,y);
	for(int i=20;~i;--i)
		if(dep[fa[x][i]]>=dep[y])
			x=fa[x][i];
	if(x==y)
		return x;
	for(int i=20;~i;--i)
		if(fa[x][i]^fa[y][i])
			x=fa[x][i],y=fa[y][i];
	return fa[x][0];
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
		f[i]=i;
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d",u+i,v+i);
		if(find(u[i])^find(v[i]))
			f[find(u[i])]=find(v[i]),add_edge(u[i],v[i]);
		else
			++c[w[i]=gen()/2];
	}
	for(int i=1;i<=n;i++)
		if(!vs[i])
			dfs(i,0);
	for(int i=1;i<=m;i++)
		if(w[i])
			g[u[i]]^=w[i],g[v[i]]^=w[i];
	for(int i=1;i<=n;i++)
		if(vs[i])
			sou(i,0);
	for(map<int,int>::iterator it=c.begin();it!=c.end();it++)
		d=gcd(d,(*it).second);
	for(int i=1;i<=n;i++)
		if(d%i==0)
			printf("%d ",i);
}