CS Academy Round 44 Check DFS

题目链接：https://csacademy.com/contest/round-44/task/check-dfs/

题目大意：给出一个无向简单连通图，顶点从1~n标号，给出一个1开头的1~n的排列，问这个排列是否可以为该图dfs一次的访问顺序。

解题思路：将给出的排列放入队列q中，pop第一个元素后dfs，对于每个节点，如果其孩子中有队首元素，那么标记该元素，队列pop，并以其为新节点搜索，搜索完毕后重复该过程；如果不包含但是孩子中有未访问的，则不可行；如果不包含但孩子没有未访问的，则返回。

注意前向星表示中边数*2！！！

代码：

 1 const int maxn = 1e5 + 5;
 2 struct Edge{
 3     int to, next;
 4 };
 5 Edge edges[maxn * 3];
 6 int tot, head[maxn * 3];
 7 queue<int> q;
 8 bool vis[maxn], flag = true;
 9 int n, m;
10 
11 void init(){
12     tot = 0;
13     flag = true;
14     memset(vis, 0, sizeof(vis));
15     memset(head, -1, sizeof(head));
16 }
17 void addEdge(int u, int v){
18     edges[tot].to = v;
19     edges[tot].next = head[u];
20     head[u] = tot++;
21 }
22 
23 void dfs(int x){
24     if(!flag) return;
25     vis[x] = 1;
26     set<int> s;
27     vector<int> vec;
28     for(int i = head[x]; i != -1; i = edges[i].next){
29         int v = edges[i].to;
30         if(!vis[v]) {
31             s.insert(v);
32             vec.push_back(v);
33         }
34     }
35     if(vec.size() == 0) return;
36     while(s.size() > 0 && q.size() > 0){
37         int tmp = q.front();
38         set<int>::iterator it = s.find(tmp);
39         if(it == s.end()){
40             for(it = s.begin(); it != s.end(); it++){
41                 if(!vis[*it]) {
42                     flag = false;
43                     return;
44                 }
45             }
46             return;
47         }
48         s.erase(it);
49         q.pop();
50         dfs(tmp);
51         if(!flag) return;
52     }
53 }
54 void solve(){
55     q.pop();
56     dfs(1);
57     if(q.size() > 0) flag = false;
58     if(flag) puts("1");
59     else puts("0");
60 }
61 int main(){
62     init();
63     scanf("%d %d", &n, &m);
64     for(int i = 1; i <= n; i++) {
65         int tmp;
66         scanf("%d", &tmp);
67         q.push(tmp);
68     }
69     for(int i = 1; i <= m; i++){
70         int u, v;
71         scanf("%d %d", &u, &v);
72         addEdge(u, v);
73         addEdge(v, u);
74     }
75     solve();
76 }

题目：

Check DFS

Time limit: 1000 ms
Memory limit: 128 MB

One of the most popular algorithms for traversing a graph is the Depth First Search (DFS).

The DFS is usually implemented as a recursive function. First we call the function for the starting node. For each call, we go through the adjacency list of the current node, and recursively call the function for the unvisited neighbours.

Notice that if we permute the adjacency lists of the nodes, the order in which we visit them might differ.

In this problem you are given a graph with

Standard input

The first line contains two integers

The second line contains a permutation of size

Each of the next

Standard output

If it's possible to visit the nodes in the given order print

Constraints and notes

$10^51≤N,M≤105$
The graph is connected, simple and undirected

Input	Output	Explanation
4 3 1 4 2 3 1 2 1 3 1 4	1	1234
4 4 1 2 4 3 1 2 2 3 3 4 2 4	1	1234
4 3 1 2 4 3 1 2 2 3 3 4	0	1234
4 6 1 4 2 3 1 2 1 3 1 4 3 2 4 2 3 4	1	1234