noip2015信息传递

最小环的思路。。。。数据也相当的水，tarjan缩点保存每个环内的点的数量，把最小的输出即可

#include<cstdio>
#include<algorithm>
const int MAXN=2e5+5;
const int INF=0x7fffffff;
struct Edge
{
	int nxt;
	int to;
}edge[MAXN<<1];
int head[MAXN];
int num;
void add(int from,int to)
{
	edge[++num].to=to;
	edge[num].nxt=head[from];
	head[from]=num;
}
int indx;
int low[MAXN];
int dfn[MAXN];
int stk[MAXN];
int color[MAXN];
int dgr[MAXN];
bool vis[MAXN];
int top;
int cnt;
void tarjan(int x)
{
	indx++;
	dfn[x]=low[x]=indx;
	stk[++top]=x;
	vis[x]=1;
	for(int i=head[x];i;i=edge[i].nxt)
	{
		int v=edge[i].to;
		if(!dfn[v])
		{
			tarjan(v);
			low[x]=std::min(low[x],low[v]);
		}
		else 
		if(vis[v])
		{
			low[x]=std::min(low[x],dfn[v]);
		}
	}
	if(dfn[x]==low[x])
	{
		cnt++;
		while(1)
		{
			int u=stk[top--];
			dgr[cnt]++;
			vis[u]=0;
			color[u]=cnt;
			if(u==x) break;
		}
	}
}
int main()
{
	int n;
	std::scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		int ti;
		std::scanf("%d",&ti);
		add(i,ti);
	}
	for(int i=1;i<=n;i++)
	{
		if(!dfn[i])
		{
			tarjan(i);
		}
	}
	int ans=INF;
	for(int i=1;i<=cnt;i++)
	{
		if(dgr[i]!=1&&dgr[i]<ans)
		{
			ans=dgr[i];
		}
	}
	printf("%d\n",ans);
	return 0;
}