[HNOI2015]菜肴制作

大意: 给定$m$个有序对$(a,b)$, 求构造一个$n$排列, 满足$m$个对中$a$均排在$b$前, 且$1$尽量靠前, 在$1$尽量靠前的前提下$2$尽量靠前,....以此类推.

答案为反向拓排的最大字典序.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif

int n, m;
int x[N], y[N], ans[N];
int deg[N];
vector<int> g[N];
priority_queue<int> q;
void work() {
	scanf("%d%d", &n, &m);
	REP(i,1,n) g[i].clear(),deg[i]=0;
	REP(i,1,m) {
		scanf("%d%d", x+i, y+i);
		g[y[i]].pb(x[i]),++deg[x[i]];
	}
	REP(i,1,n) if (!deg[i]) q.push(i);
	int cnt = 0;
	while (q.size()) {
		int u = q.top(); q.pop();
		ans[++cnt] = u;
		for (int v:g[u]) {
			if (!--deg[v]) q.push(v);
		}
	}
	if (cnt!=n) return puts("Impossible!"),void();
	PER(i,1,n) printf("%d ", ans[i]);hr;
}

int main() {
	int t;
	scanf("%d", &t);
	while (t--) work();
}