[2020.10.28 SSL Dice T6] [Rakutani P2341] POPULAR [dfs and a little shearing]

$link$

analysis:

The data is relatively watery ,
you can directly $DFS$ through the notemark + Analyzingto
this$dfs$ can't live on Luogu, the
positive solution is$taijan$ shrink point or$kosaraju\; to$ findconnected components
andsearch + pruningcan also pass
$n^2$ more than a million violent search grind scalar operators
cattle node asthe reverse side was builtfor a cow ran$DFS$
judgment ability to non-arrival达其otherownership奶牛Shikakisaki pressurized$SmalltoothplaceorRiof$ pruning immediately Allowed
还要开个$O\left(2\right)$ optimization

CODE：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int N=50005;
int n,m,vis[N],ans[N],head[N],tot,cnt;
struct node{
    
    
	int to,next;
}a[N];
void add(int x,int y)
{
    
    
	a[++tot]=(node){
    
    y,head[x]};  //邻接表
	head[x]=tot;
}
void dfs(int dep)
{
    
    
	vis[dep]=1;
	for(int i=head[dep];i;i=a[i].next)
	{
    
    
		int qwq=a[i].to;
		if(!vis[qwq])  //未访问
		{
    
    
			vis[qwq]=1;
			ans[qwq]++;
			dfs(qwq);
		}
	}
}
int main(){
    
    
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	{
    
    
		int x,y;
		scanf("%d%d",&x,&y);
		add(x,y);
	}
	for(int i=1;i<=n;i++)
	{
    
    
		memset(vis,0,sizeof(vis));
		dfs(i);
	}
	for(int i=1;i<=n;i++)
		if(ans[i]>=n-1) cnt++;
	printf("%d",cnt);
	
	return 0;
} 

Rakutani AC's dfs:

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#define reg register
using namespace std;
const int N=10005;
int n,m,vis[N],all[N],lqm,a[N],cnt;
vector<int> G[N]; 
void dfs(int dep)  //dfs
{
    
    
	if(lqm==1) return;  //存在ans
	for(reg int i=0;i<a[dep];i++)  
		if(!vis[G[dep][i]])  //未访问
		{
    
    
			vis[G[dep][i]]=1;
			if(all[G[dep][i]]==1)
			{
    
    
				lqm=all[i];  //能否到达全部
				return;
			}
			dfs(G[dep][i]);		
		}
}
int main(){
    
    
	scanf("%d%d",&n,&m);
	for(reg int i=0;i<m;i++)
	{
    
    
		int x,y;
		scanf("%d%d",&x,&y);
		vis[x]=1;vis[y]=1;
		G[y].push_back(x);
	}
	bool f=0;
	for(reg int i=1;i<=n;i++)
		if(vis[i]==0)
		{
    
    
			f=1;break;  //牛没有在图中
		}
	if(f){
    
    
		printf("0");
		return 0;
	}
	for(reg int i=1;i<=n;i++)
	{
    
    
		sort(G[i].begin(),G[i].end());
		a[i]=unique(G[i].begin(),G[i].end())-G[i].begin();  //排序，去重
	}
	memset(all,-1,sizeof(all));
	for(reg int i=1;i<=n;i++)
	{
    
    
		if(all[i]==0) continue;  //无法到达全部
		lqm=-1;
		memset(vis,0,sizeof(vis));
		vis[i]=1;dfs(i);
		if(lqm==1)
		{
    
    
			all[i]=1;
			cnt++;
		} 
		else if(lqm==-1)
		{
    
    
			bool f=0;
			for(reg int j=1;j<=n;j++)
				if(vis[j]==0)
				{
    
    
					f=1;
					break;
				}
			if(f==0)
			{
    
    
				all[i]=1;
				cnt++;
			}else
				for(reg int j=1;j<=n;j++)
					if(vis[j]) all[j]=0;  //剪枝
				//如果此牛不能到达全部 则此牛到达的全部牛无法到达全部牛
		}
	}
	printf("%d",cnt);
	
	return 0;
}