逃生（拓扑排序）

题目链接
题意：n个人编号1-n，有m种限制关系，u、v表示u必须在v前面，编号小尽量放前面。给出编号顺序。
解法：建立一个反图，跑一边字典序最大的拓扑排序，最后再把这个排序倒过来就是答案了。
参考博客

#include<bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
#include <stdlib.h>
#include <time.h>
using namespace std;
typedef long long ll ;
#define int ll
#define mod 10000
#define gcd __gcd
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j)  for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
//int lcm(int a , int b){return a*b/gcd(a,b);}
//ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define INF  0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define pb push_back
#define mp make_pair
#define all(v) v.begin(),v.end()
#define size(v) (int)(v.size())
#define cin(x) scanf("%lld" , &x);
const int N = 11;
const int maxn = 3e4+9;
const double esp = 1e-6;
int in[maxn] , n, m;
vector<int>g[maxn];


void init(){
    ME(in , 0);
    rep(i , 1 , n){
        g[i].clear();
    }
}
void tuopu(){
    vector<int>ans;
    priority_queue<int>q;
    rep(i , 1 , n){
        if(!in[i]){
            q.push(i);
        }
    }
    while(!q.empty()){
        int u = q.top();q.pop();
        ans.pb(u);
        for(auto &v : g[u]){
            in[v]--;
            if(!in[v]){
                q.push(v);
            }
        }
    }
    for(int i = size(ans)-1 ; i > 0 ; i--)
        cout << ans[i] << " ";
    cout << ans[0] << endl;
}

void solve(){
    scanf("%lld%lld" , &n , &m);
    init();
    rep(i , 1 , m){
        int u , v;
        scanf("%lld%lld" , &u , &v);
        g[v].pb(u);
        in[u]++;
    }
    tuopu();
}
signed main()
{
    //ios::sync_with_stdio(false);
    int t ;
    scanf("%lld" , &t);
    while(t--)
        solve();
}