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
1 /**
2 * Codeforces
3 * Problem#825E
4 * Accepted
5 * Time: 46ms
6 * Memory: 5300k
7 */
8 #include <bits/stdc++.h>
9 #ifndef WIN32
10 #define Auto "%lld"
11 #else
12 #define Auto "%I64d"
13 #endif
14 using namespace std;
15 typedef bool boolean;
16 const signed int inf = (signed)((1u << 31) - 1);
17 const double eps = 1e-6;
18 const int binary_limit = 128;
19 #define smin(a, b) a = min(a, b)
20 #define smax(a, b) a = max(a, b)
21 #define max3(a, b, c) max(a, max(b, c))
22 #define min3(a, b, c) min(a, min(b, c))
23 template<typename T>
24 inline boolean readInteger(T& u){
25 char x;
26 int aFlag = 1;
27 while(!isdigit((x = getchar())) && x != ‘-‘ && x != -1);
28 if(x == -1) {
29 ungetc(x, stdin);
30 return false;
31 }
32 if(x == ‘-‘){
33 x = getchar();
34 aFlag = -1;
35 }
36 for(u = x - ‘0‘; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - ‘0‘);
37 ungetc(x, stdin);
38 u *= aFlag;
39 return true;
40 }
41
42 ///map template starts
43 typedef class Edge{
44 public:
45 int end;
46 int next;
47 Edge(const int end = 0, const int next = -1):end(end), next(next){}
48 }Edge;
49
50 typedef class MapManager{
51 public:
52 int ce;
53 int *h;
54 vector<Edge> edge;
55 MapManager(){}
56 MapManager(int points):ce(0){
57 h = new int[(const int)(points + 1)];
58 memset(h, -1, sizeof(int) * (points + 1));
59 }
60 inline void addEdge(int from, int end){
61 edge.push_back(Edge(end, h[from]));
62 h[from] = ce++;
63 }
64 inline void addDoubleEdge(int from, int end){
65 addEdge(from, end);
66 addEdge(end, from);
67 }
68 Edge& operator [] (int pos) {
69 return edge[pos];
70 }
71 inline void clear() {
72 edge.clear();
73 delete[] h;
74 }
75 }MapManager;
76 #define m_begin(g, i) (g).h[(i)]
77 #define m_endpos -1
78 ///map template ends
79
80 int n, m;
81 MapManager g;
82 int* dag;
83 int* dep;
84
85 inline boolean init() {
86 if(!readInteger(n)) return false;
87 readInteger(m);
88 g = MapManager(n);
89 dag = new int[(n + 1)];
90 dep = new int[(n + 1)];
91 memset(dag, 0, sizeof(int) * (n + 1));
92 for(int i = 1, a, b; i <= m; i++) {
93 readInteger(a);
94 readInteger(b);
95 g.addEdge(b, a);
96 dag[a]++;
97 }
98 return true;
99 }
100
101 priority_queue<int> que;
102 inline void topu() {
103 for(int i = 1; i <= n; i++)
104 if(!dag[i]) {
105 que.push(i);
106 }
107 int cnt = 0;
108 while(!que.empty()) {
109 int e = que.top();
110 dep[e] = cnt++;
111 que.pop();
112 for(int i = m_begin(g, e); i != m_endpos; i = g[i].next) {
113 int& eu = g[i].end;
114 dag[eu]--;
115 if(!dag[eu])
116 que.push(eu);
117 }
118 }
119 }
120
121 inline void solve() {
122 topu();
123 for(int i = 1; i <= n; i++)
124 printf("%d ", n - dep[i]);
125 }
126
127 int main() {
128 init();
129 solve();
130 return 0;
131 }
时间： 2024-10-12 15:47:32