拓扑序
作者:
amagi
,
2023-07-14 13:14:48
,
所有人可见
,
阅读 136
using namespace std;
const int N = 1e5 + 100;
int ne[N],e[N],idx;
int h[N];
int d[N];
int n,m;
int q[N];
void add(int a,int b){
e[idx] = b;
ne[idx] = h[a];
h[a] = idx ++;
}
bool ss(){
int hh = 1,zz = 0;
for(int i = 1; i <= n; i ++){
if(d[i] == 0) q[++ zz] = i;
}
while(hh <= zz){
int z = q[hh ++];
for(int i = h[z]; i != -1; i = ne[i]){
int j = e[i];
if(--d[j] == 0) q[++ zz] = j;
}
}
return zz == n;
}
int main(){
cin >> n >> m;
memset(h,-1,sizeof h);
for(int i = 1; i <= m; i ++){
int a,b;
cin >> a >> b;
add(a,b);
d[b] ++;
}
if(ss()){
for(int i = 1; i <= n; i ++) cout << q[i] << " ";
cout << endl;
}
else cout << -1 << endl;
return 0;
}