codeforces 412D. Giving Awards （dfs）

​​http://codeforces.com/problemset/problem/412/D​​

Description

The employees of the R1 company often spend time together: they watch football, they go camping, they solve contests. So, it's no big deal that sometimes someone pays for someone else.

Today is the day of giving out money rewards. The R1 company CEO will invite employees into his office one by one, rewarding each one for the hard work this month. The CEO knows who owes money to whom. And he also understands that if he invites person x to his office for a reward, and then immediately invite person y, who has lent some money to person x, then they can meet. Of course, in such a situation, the joy of person x

However, there are a lot of employees in the company, and the CEO doesn't have a lot of time. Therefore, the task has been assigned to you. Given the debt relationships between all the employees, determine in which order they should be invited to the office of the R1 company CEO, or determine that the described order does not exist.

Input

The first line contains space-separated integers n and m

— the number of employees in R1 and the number of debt relations. Each of the following m lines contains two space-separated integers a_i, b_i(1 ≤ a_i, b_i ≤ n; a_i ≠ b_i), these integers indicate that the person number a_i owes money to a person a number b_i. Assume that all the employees are numbered from 1 to n.

It is guaranteed that each pair of people p, q is mentioned in the input data at most once. In particular, the input data will not contain pairs p, q and q, p

Output

Print -1 if the described order does not exist. Otherwise, print the permutation of n

If there are multiple correct orders, you are allowed to print any of them.

Sample Input

Input

2 11 2

Output

2 1

Input

3 31 2
2 3
3 1

Output

2 1 3

分析：求出一个排列，它里面不能出现给定关系的顺序，比如例子中的1 2; 2 3; 3 1 对应output没有任何直接的1 2或 2 3或3 1

应用dfs查询给定的关系，然后逆序输出，但是我想不到输出-1的情况，也的确没有嘛。果然A了。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N=3e4+10,M=1e5+10;
int p[N];
bool vis[N];
struct node{
    int u,v,next;
}edge[M];
int head[M],top,sum,n,m;
void add(int a,int b){
     edge[top].u=a;
     edge[top].v=b;
     edge[top].next=head[a];
     head[a]=top++;
}
void dfs(int a){
     if(vis[a]) return ;
     vis[a]=1;
     for(int i=head[a];i>-1;i=edge[i].next){
         dfs(edge[i].v);
     }
     if(sum<n-1)printf("%d ",a);
     else printf("%d\n",a);
     sum++;
}
int main()
{
    //freopen("cin.txt","r",stdin);
    int a,b;
    while(cin>>n>>m){
        top=sum=0;
        memset(vis,0,sizeof(vis));
        memset(head,-1,sizeof(head));
        for(int i=0;i<m;i++){
            scanf("%d%d",&a,&b);
            add(a,b);
        }
        for(int i=1;i<=n;i++){
            dfs(i);
        }
    }
    return 0;
}