codeforces 412D. Giving Awards （dfs）

最新推荐文章于 2020-06-04 00:30:09 发布

theArcticOcean

最新推荐文章于 2020-06-04 00:30:09 发布

阅读量796

点赞数

CC 4.0 BY-SA版权

分类专栏： algorithm_搜索文章标签：递归

本文链接：https://blog.youkuaiyun.com/thearcticocean/article/details/50042735

algorithm_搜索专栏收录该内容

16 篇文章

订阅专栏

该博客讨论了Codeforces 412D问题，即如何确定一个公司员工的奖励顺序，使得在CEO依次邀请员工时，不会出现欠债者紧接着被债主邀请的情况。题目描述了输入输出格式，并提到了可以使用DFS解决，通过查询债务关系并逆序输出。虽然存在多种正确顺序，但博主未能想到需要输出-1的情况，即不存在满足条件的顺序，不过在实际情况下问题有解。

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 from his brand new money reward will be much less. Therefore, the R1 CEO decided to invite the staff in such an order that the described situation will not happen for any pair of employees invited one after another.

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 $\text{[math]}$ — 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 simultaneously.

Output

Print -1 if the described order does not exist. Otherwise, print the permutation of n distinct integers. The first number should denote the number of the person who goes to the CEO office first, the second number denote the person who goes second and so on.

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

Sample Input

Input

2 1
1 2

Output

2 1

Input

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;
}