1122. Hamiltonian Cycle (25)

ゝ一世哀愁。 2022-05-28 12:36 212阅读 0赞

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 ... Vn

where n is the number of vertices in the list, and Vi'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

--------------------

题目大意：

给定一个无向图，以及遍历序列，判断其是否为哈密顿回路。

考虑哈密顿回路的必要条件：

*  序列首尾相同
 *  序列长度 = 顶点个数 + 1
 *  除首尾外，不会再有相同的顶点。
 *  序列中相邻的两个顶点要求有边相连。

满足这些条件可以判断这是一条哈密顿回路。

代码：

#include<stdio.h>
    #include<vector>
    #include<string.h>
    using namespace std;
    int Map[210][210];
    vector<int> v;
    int n;
    int num[210];
    int judge()
    {
        int i,j;
        if(v[0]!=v[v.size()-1])
        {
            return 0;
        }
        else
        {
            for(i=0;i<v.size()-1;i++)
            {
                if(Map[v[i]][v[i+1]]!=1)
                {
                    break;
                }
            }
            if(i<v.size()-1)
            {
                return 0;
            }
            for(i=1;i<=n;i++)
            {
                if(num[i]>2)
                    break;
                else if(num[i]==2&&i!=v[0])
                    break;
                else if(num[i]<1)
                    break;
            }
            if(i<=n)
            {
                return 0;
            }
        }
        return 1;
    }
    int main()
    {
        int i,j,m,k,t,x,y,node;
        scanf("%d %d",&n,&m);
        for(i=0;i<m;i++)
        {
            scanf("%d %d",&x,&y);
            Map[x][y]=Map[y][x]=1;
        }
        scanf("%d",&k);
        for(i=0;i<k;i++)
        {
            memset(num,0,sizeof(num));
            scanf("%d",&t);
            v.clear();
            for(j=0;j<t;j++)
            {
                scanf("%d",&node);
                v.push_back(node);
                num[node]++;
            }
            if(t!=n+1)
            {
                printf("NO\n");
            }
            else
            {
               if(judge())
               {
                   printf("YES\n");
               }
               else
               {
                   printf("NO\n");
               }
            }
        }
        return 0;
    }