ACM ICPC 2015 Moscow Subregional Russia, Moscow, Dolgoprudny, October, 18, 2015 K. King’s Rout

K. King's Rout

time limit per test

4 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

The great rout will be held this evening in the palace of his majesty Nassah II, the king of Occorom. There are n guests invited. While they are preparing evening dresses and collecting fresh rumors to talk about, the chief valet of the palace has a tricky task to solve: choose the right order for persons to arrive to the palace.

Guests always arrive one by one, that is, no two guests may arrive at the same moment of time. Due to the court etiquette, there are some limitations on the order of the arrival. For example, a notable landlord should arrive later than all his vassals, but should be earlier than his wives. After reading "Etiquette guide for dummies" the valet found out m order conditions to be satisfied. Each of them has a form: ai must come before bi. Rules are so complicated that some conditions may appear in the list two or more times.

So far the problem seems to be easy and familiar. But some guests (actually, all of them) tried to bribe valet to allow them arrive before others. So valet sorted guests according to their payoffs amounts and decided that guest number 1 should arrive as early as possible (without violating etiquette rules), among all such options valet chooses the one with the guest number 2 arriving as early as possible, and so on. All payoffs were different, so valet has no problem in selecting guests priority.

Help valet to find the best possible schedule. Guests already have numbers in valet's private list of priority, so you will not know bribes amounts and will not be accused in complicity in corruption.

Input

The first line of the input contains two integers n and m (1 ≤ n ≤ 200 000, 0 ≤ m ≤ 400 000) — the number of guests invited and the number of order conditions respectively.

Next m lines describe the conditions, each of them containing a single pair ai, bi (1 ≤ ai, bi ≤ n). That means the guest ai is required to come earlier than the guest bi.

Output

Print n different integers from 1 to n to describe the best possible order (according to valet's understanding) for guests to arrive. It is guaranteed that at least one valid order exists.

Sample test(s)

input

3 1
3 1

output

3 1 2 

input

5 6
2 1
5 2
4 1
5 4
3 1
5 3

output

5 2 3 4 1 

Note

In the first sample all the permutations where guest number 1 comes after guest number 3 are acceptable according to etiquette. As the valet wants the guest number 1 to come as early as possible he puts him on the second slot and the guest number 3 on the first slot. There is only one slot remaining for the guest number 2.

题意：给出N个点，M条有向边，求字典序最小的topo排序。

分析：注意是字典序最小。

有优先队列小心处理下即可。

小心被坑

 /**
 Create By yzx - stupidboy
 */
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <cmath>
 #include <deque>
 #include <vector>
 #include <queue>
 #include <iostream>
 #include <algorithm>
 #include <map>
 #include <set>
 #include <ctime>
 #include <iomanip>
 using namespace std;
 typedef long long LL;
 typedef double DB;
 #define MIT (2147483647)
 #define INF (1000000001)
 #define MLL (1000000000000000001LL)
 #define sz(x) ((int) (x).size())
 #define clr(x, y) memset(x, y, sizeof(x))
 #define puf push_front
 #define pub push_back
 #define pof pop_front
 #define pob pop_back
 #define ft first
 #define sd second
 #define mk make_pair

 inline int Getint()
 {
     int Ret = ;
     char Ch = ' ';
     bool Flag = ;
     while(!(Ch >= '' && Ch <= ''))
     {
         if(Ch == '-') Flag ^= ;
         Ch = getchar();
     }
     while(Ch >= '' && Ch <= '')
     {
         Ret = Ret *  + Ch - '';
         Ch = getchar();
     }
     return Flag ? -Ret : Ret;
 }

 const int N = , M = ;
 int n, m;
 int cnt[N];
 vector<int> fa[N];
 priority_queue<int> ranks;
 vector<int> ans;

 inline void Input()
 {
     scanf("%d%d", &n, &m);
     for(int i = ; i < m; i++)
     {
         int x, y;
         scanf("%d%d", &x, &y);
         fa[y].pub(x), cnt[x]++;
     }
 }

 inline void Solve()
 {
     for(int i = ; i <= n; i++)
         if(!cnt[i]) ranks.push(i);
     for(int index = n; index >= ; index--)
     {
         int u = ranks.top();
         ranks.pop();
         ans.pub(u);
         int length = sz(fa[u]);
         for(int i = , v; i < length; i++)
             if(!(--cnt[v = fa[u][i]]))
                 ranks.push(v);
     }

     int length = sz(ans);
     for(int i = length - ; i > ; i--)
         printf("%d ", ans[i]);
     printf("%d\n", ans[]);
 }

 int main()
 {
     Input();
     Solve();
     return ;
 }