题目：https://www.luogu.org/problemnew/show/P2661

此题表面上是一道求强连通分量的题，实际上很简单无需算法

因为出度为1，所以直接循环就行了。。。

搞了好久恍然大悟

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxSize=200000,maxValue=0x7fffffff;
int n;
int a[maxSize+5],vis[maxSize+5],ru[maxSize+5],dis[maxSize+5],ans;

void dfs1(int x)
{
	int i;
	
	printf("%d %d\n",x,dis[x]);
	if (vis[a[x]]==1)
	{
		ans=min(ans,dis[x]-dis[a[x]]+1);
		return ;
	}
	vis[a[x]]=1;
	dis[a[x]]=dis[x]+1;
	dfs1(a[x]);
	dis[a[x]]=0;
}

void dfs(int x)
{
	int i;
	
	while (vis[a[x]]==0)
	{
		vis[a[x]]=1;
		dis[a[x]]=dis[x]+1;
		x=a[x];
	}
	ans=min(ans,dis[x]-dis[a[x]]+1);
}

int main()
{
	int i,x;
	
	freopen("a.txt","r",stdin);
	freopen("b.txt","w",stdout);
	scanf("%d",&n);
	
	memset(vis,0,sizeof(vis));
	memset(ru,0,sizeof(ru));
	for (i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
		ru[a[i]]++;
	}
	for (i=1;i<=n;i++)
	{
		x=i;
		while (ru[x]==0 && vis[x]==0)
		{
			vis[x]=1;
			ru[a[x]]--;			
			x=a[x];
		}
	}
	ans=maxValue;
	for (i=1;i<=n;i++)
		if (vis[i]==0)
			dfs(i);
	printf("%d\n",ans);
	
	return 0;
}