PAT甲级——A1126 Eulerian Path【30】

In graph theory, an Eulerian path is a path in a graph which visits every edge exactly once. Similarly, an Eulerian circuit is an Eulerian path which starts and ends on the same vertex. They were first discussed by Leonhard Euler while solving the famous Seven Bridges of Konigsberg problem in 1736. It has been proven that connected graphs with all vertices of even degree have an Eulerian circuit, and such graphs are called Eulerian. If there are exactly two vertices of odd degree, all Eulerian paths start at one of them and end at the other. A graph that has an Eulerian path but not an Eulerian circuit is called semi-Eulerian. (Cited from https://en.wikipedia.org/wiki/Eulerian_path)

Given an undirected graph, you are supposed to tell if it is Eulerian, semi-Eulerian, or non-Eulerian.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 2 numbers N (≤ 500), and M, which are the total number of vertices, and the number of edges, respectively. Then M lines follow, each describes an edge by giving the two ends of the edge (the vertices are numbered from 1 to N).

Output Specification:

For each test case, first print in a line the degrees of the vertices in ascending order of their indices. Then in the next line print your conclusion about the graph -- either Eulerian, Semi-Eulerian, or Non-Eulerian. Note that all the numbers in the first line must be separated by exactly 1 space, and there must be no extra space at the beginning or the end of the line.

Sample Input 1:

7 12
5 7
1 2
1 3
2 3
2 4
3 4
5 2
7 6
6 3
4 5
6 4
5 6

Sample Output 1:

2 4 4 4 4 4 2
Eulerian

Sample Input 2:

6 10
1 2
1 3
2 3
2 4
3 4
5 2
6 3
4 5
6 4
5 6

Sample Output 2:

2 4 4 4 3 3
Semi-Eulerian

Sample Input 3:

5 8
1 2
2 5
5 4
4 1
1 3
3 2
3 4
5 3

Sample Output 3:

3 3 4 3 3
Non-Eulerian

题⽬⼤意：如果⼀个连通图的所有结点的度都是偶数，那么它就是Eulerian，如果除了两个结点的度是

奇数其他都是偶数，那么它就是Semi-Eulerian，否则就是Non-Eulerian～

分析：⽤邻接表存储图，判断每个结点的度【也就是每个结点i的v[i].size()】是多少即可得到最终结果

～注意：图必须是连通图，所以要⽤⼀个深搜判断⼀下连通性，从结点1开始深搜，如果最后发现统计

的连通结点个数cnt != n说明是不是连通图，要输出Non-Eulerian～

#include <iostream>
using namespace std;
int path[505] = { 0 }, graph[505][505];
int n, m, a, b, odd = 0, num = 0;
bool visit[505] = { false };
void DFS(int s)
{
    visit[s] = true;
    num++;
    for (int i = 1; i <= n; ++i)
        if (graph[s][i] == 1 && visit[i] == false)            
            DFS(i);
}
int main()
{    
    cin >> n >> m;    
    while (m--)
    {
        cin >> a >> b;
        path[a]++;
        path[b]++;
        graph[a][b] = graph[b][a] = 1;
    }
    for (int i = 1; i <= n; ++i)
    {
        if (path[i] % 2 == 1)
            odd++;
        cout << (i == 1 ? "" : " ") << path[i];
    }
    DFS(1);//判断是不是连通图
    if (num == n && odd == 0)
        cout << endl << "Eulerian" << endl;
    else if (num == n && odd == 2)
        cout << endl << "Semi-Eulerian" << endl;
    else
        cout << endl << "Non-Eulerian" << endl;
    return 0;
}