1122 Hamiltonian Cycle (25 分)

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<N≤200), 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 K which is the number of queries, followed by K lines of queries, each in the format:

n V1 V2 … 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:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

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:

1 2 3 4 5 6

YES NO NO NO YES NO

作者: CHEN, Yue

单位: 浙江大学

时间限制: 300 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

给出一个图，要求判断一个环是否是Hamiltonian cycle。题目对这个定义说得挺不清楚的。其实就是必须是简单环（除起点终点不能重复）而且图中所有点都要访问一次。

代码

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 大专栏  1122 Hamiltonian Cycle (25 分)>54 55 56 57 58 59 60

#include <vector> #include <stack> #include <algorithm> #include <deque> #include <queue> #include <map> #include <set> #include <cstring> #include <cmath> #include <sstream> using namespace std; int n,m,k; int visited[202]; int g[202][202]; int () { cin>>n>>m; int v1,v2; for(int i=0;i<m;i++){ cin>>v1>>v2; g[v1][v2]=1; g[v2][v1]=1; } cin>>k; int c; for(int i=0;i<k;i++){ cin>>c; int first,cur,pre,t; bool iscycle=true; fill(visited,visited+202,0); for(int j=0;j<c;j++){ cin>>t; visited[t]++; if(j==0){ first=t; pre=t; } else{ cur=t; if(g[pre][cur]==0) iscycle=false; pre=cur; } } for(int i=1;i<=n;i++){ if(i!=first&&visited[i]!=1) iscycle=false; if(i==first&&visited[i]!=2) iscycle=false; } if(cur!=first) iscycle=false; iscycle?cout<<"YES":cout<<"NO"; cout<<endl; } return 0; }