1122 Hamiltonian Cycle (25 分)
The "Hamilton cycle problem" is to find a simple cycle that contains every vertex in a graph. Such a cycle is called a "Hamiltonian cycle".
In this problem, you are supposed to tell if a given cycle is a Hamiltonian cycle.
Input Specification:
Each input file contains one test case. For each case, the first line contains 2 positive integers N (2), the number of vertices, and M, the number of edges in an undirected graph. Then M lines follow, each describes an edge in the format Vertex1 Vertex2, where the vertices are numbered from 1 to N. The next line gives a positive integer Kwhich is the number of queries, followed by K lines of queries, each in the format:
n V?1?? V?2?? ... V?n??
where n is the number of vertices in the list, and V?i??‘s are the vertices on a path.
Output Specification:
For each query, print in a line YES if the path does form a Hamiltonian cycle, or NO if not.
Sample Input:
6 10
6 2
3 4
1 5
2 5
3 1
4 1
1 6
6 3
1 2
4 5
6
7 5 1 4 3 6 2 5
6 5 1 4 3 6 2
9 6 2 1 6 3 4 5 2 6
4 1 2 5 1
7 6 1 3 4 5 2 6
7 6 1 2 5 4 3 1
Sample Output:
YES
NO
NO
NO
YES
NO
就多判断几下了，感觉都不涉及图的知识
1 #include <bits/stdc++.h>
2 using namespace std;
3 int n, m, k, p;
4 vector<int> vt;
5 int v[500][500];
6 int main(){
7 cin >> n >> m;
8 int x, y;
9 for(int i = 0; i < m; i++){
10 cin >> x >> y;
11 v[x][y] = 1;
12 v[y][x] = 1;
13 }
14 cin >> k;
15 while(k--){
16 cin >> p;
17 vt.clear();
18 for(int i = 0; i < p; i++){
19 cin >> x;
20 vt.push_back(x);
21 }
22 set<int> s;
23 if(p == n+1){
24 if(vt[0] == vt[vt.size()-1]){
25 bool flag = true;
26 for(int i = 1; i < vt.size(); i++){
27 if(v[vt[i-1]][vt[i]] == 0){
28 flag = false;
29 break;
30 }
31 s.insert(vt[i]);
32 }
33 if(flag && s.size() == n){
34 cout << "YES" << endl;
35 }else{
36 cout <<"NO"<<endl;
37 }
38 }else{
39 cout << "NO" << endl;
40 }
41 }else{
42 cout << "NO" << endl;
43 }
44 }
45
46 return 0;
47 }
原文地址：https://www.cnblogs.com/zllwxm123/p/11326501.html