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:

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

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

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大专栏  1122 Hamiltonian Cycle (25 分)>54
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#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;
}