从一个点正向访问的所有的点,也能从这个点反向访问到所有的点,则yes,否则no;
#include<iostream>
#include<cstring>
#include<vector>
#define maxn 100000+5
using namespace std;
int n,m;
vector<int>zh[maxn];
vector<int>fa[maxn];
int visit[10000+5];
int sum1,sum2;
void dfsz(int x)
{
visit[x]=1;
for(int i=0;i<zh[x].size();i++)
{
if(!visit[zh[x][i]]) dfsz(zh[x][i]),sum1++;
}
}
void dfsf(int x)
{
visit[x]=1;
for(int i=0;i<fa[x].size();i++)
{
if(!visit[fa[x][i]]) dfsf(fa[x][i]),sum2++;
}
}
int main()
{
while(cin>>n>>m)
{
if(!n&&!m) break;
for(int i=0;i<maxn;i++) zh[i].clear(),fa[i].clear();
while(m--)
{
int x,y;
cin>>x>>y;
zh[x].push_back(y);
fa[y].push_back(x);
}
sum1=1;sum2=1;
fill(visit,visit+n+1,0);
dfsz(1);
fill(visit,visit+n+1,0);
dfsf(1);
if(sum1==n&&sum2==n) cout<<"Yes"<<endl;
else cout<<"No"<<endl;
}
return 0;
}