Codeforces 825E Minimal Labels - 拓扑排序 - 贪心

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 33 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

　　题目大意 给定一个有向无环图，用1~n为所有顶点标号，每个顶点的标号互不相同，如果有有一条边从v连向u，则v的标号应比u小，输出字典序最小的标号方案。

　　poj有一道一样的题，题解请戳这里。

Code

/**
* Codeforces
* Problem#825E
* Accepted
* Time: 46ms
* Memory: 5300k
*/
#include <bits/stdc++.h>
#ifndef WIN32
#define Auto "%lld"
#else
#define Auto "%I64d"
#endif
using namespace std;
typedef bool boolean;
const signed int inf = (signed)((1u << 31) - 1);
const double eps = 1e-6;
const int binary_limit = 128;
#define smin(a, b) a = min(a, b)
#define smax(a, b) a = max(a, b)
#define max3(a, b, c) max(a, max(b, c))
#define min3(a, b, c) min(a, min(b, c))
template<typename T>
inline boolean readInteger(T& u){
char x;
int aFlag = 1;
while(!isdigit((x = getchar())) && x != '-' && x != -1);
if(x == -1) {
ungetc(x, stdin);
return false;
}
if(x == '-'){
x = getchar();
aFlag = -1;
}
for(u = x - '0'; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - '0');
ungetc(x, stdin);
u *= aFlag;
return true;
}

///map template starts
typedef class Edge{
public:
int end;
int next;
Edge(const int end = 0, const int next = -1):end(end), next(next){}
}Edge;

typedef class MapManager{
public:
int ce;
int *h;
vector<Edge> edge;
MapManager(){}
MapManager(int points):ce(0){
h = new int[(const int)(points + 1)];
memset(h, -1, sizeof(int) * (points + 1));
}
inline void addEdge(int from, int end){
edge.push_back(Edge(end, h[from]));
h[from] = ce++;
}
inline void addDoubleEdge(int from, int end){
addEdge(from, end);
addEdge(end, from);
}
Edge& operator [] (int pos) {
return edge[pos];
}
inline void clear() {
edge.clear();
delete[] h;
}
}MapManager;
#define m_begin(g, i) (g).h[(i)]
#define m_endpos -1
///map template ends

int n, m;
MapManager g;
int* dag;
int* dep;

inline boolean init() {
if(!readInteger(n))    return false;
readInteger(m);
g = MapManager(n);
dag = new int[(n + 1)];
dep = new int[(n + 1)];
memset(dag, 0, sizeof(int) * (n + 1));
for(int i = 1, a, b; i <= m; i++) {
readInteger(a);
readInteger(b);
g.addEdge(b, a);
dag[a]++;
}
return true;
}

priority_queue<int> que;
inline void topu() {
for(int i = 1; i <= n; i++)
if(!dag[i]) {
que.push(i);
}
int cnt = 0;
while(!que.empty()) {
int e = que.top();
dep[e] = cnt++;
que.pop();
for(int i = m_begin(g, e); i != m_endpos; i = g[i].next) {
int& eu = g[i].end;
dag[eu]--;
if(!dag[eu])
que.push(eu);
}
}
}

inline void solve() {
topu();
for(int i = 1; i <= n; i++)
printf("%d ", n - dep[i]);
}

int main() {
init();
solve();
return 0;
}