E. Minimal Labels
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given a directed acyclic graph with n vertices and m edges. There are no self-loops or multiple edges between any pair of vertices. Graph can be disconnected.
You should assign labels to all vertices in such a way that:
- Labels form a valid permutation of length n — an integer sequence such that each integer from 1 to n appears exactly once in it.
- If there exists an edge from vertex v to vertex u then labelv should be smaller than labelu.
- Permutation should be lexicographically smallest among all suitable.
Find such sequence of labels to satisfy all the conditions.
Input
The first line contains two integer numbers n, m (2 ≤ n ≤ 105, 1 ≤ m ≤ 105).
Next m lines contain two integer numbers v and u (1 ≤ v, u ≤ n, v ≠ u) — edges of the graph. Edges are directed, graph doesn't contain loops or multiple edges.
Output
Print n numbers — lexicographically smallest correct permutation of labels of vertices.
Examples
input
3 3
1 2
1 3
3 2
output
1 3 2
input
4 5
3 1
4 1
2 3
3 4
2 4
output
4 1 2 3
input
5 4
3 1
2 1
2 3
4 5
output
3 1 2 4 5
题意：给你n个点，m条边的有向无环图，一条边u->v表示u的权值小于v的权值；求字典序最小的方案；
思路：反向建图，使得大的点的权值更大；类似与hdu 4857；
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define LL long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x) cout<<"bug"<<x<<endl;
const int N=1e5+,M=2e6+,inf=1e9+;
const LL INF=1e18+,mod=1e9+; vector<int>edge[N];
int du[N],ans[N];
priority_queue<int>q;
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=;i<=m;i++)
{
int u,v;
scanf("%d%d",&u,&v);
edge[v].push_back(u);
du[u]++;
}
for(int i=;i<=n;i++)
if(!du[i])q.push(i);
int s=n;
while(!q.empty())
{
int x=q.top();
q.pop();
ans[x]=s--;
for(int i=;i<edge[x].size();i++)
{
int v=edge[x][i];
du[v]--;
if(!du[v])q.push(v);
}
}
for(int i=;i<=n;i++)
printf("%d ",ans[i]);
return ;
}