1126 Eulerian Path

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

题意：

　　一笔画问题（Eulerian graph）是图论中一个著名的问题。一笔画问题起源于柯尼斯堡七桥问题。数学家欧拉在他1736年发表的论文《柯尼斯堡的七桥》中不仅解决了七桥问题，也提出了一笔画定理，顺带解决了一笔画问题。

　　定理

　　　　连通的无向图 G 有欧拉路径的充要条件是：G 中奇顶点（连接的边数量为奇数的顶点）的数目等于0或者2。

　　　　连通的无向图 G 是欧拉环（存在欧拉回路）的充要条件是：G 中每个顶点的度都是偶数

思路：

　　根据每个顶点的度的奇偶性来判断是不是Eulerian graph，可能存在不连通的图因此需要DFS判断他是不是连通。

Code:

#include <bits/stdc++.h>

using namespace std;

vector<int> grap[505];
set<int> s;

void DFS(int index) {
    if (s.find(index) != s.end()) return;
    s.insert(index);
    for (int g : grap[index]) {
        DFS(g);
    }
}

int main() {
    int n, m, v1, v2;
    cin >> n >> m;
    for (int i = 0; i < m; ++i) {
        cin >> v1 >> v2;
        grap[v1].push_back(v2);
        grap[v2].push_back(v1);
    }
    int even = 0, odd = 0;
    cout << grap[1].size();
    for (int i = 1; i <= n; ++i) {
        if (i > 1) cout << " " << grap[i].size();
        if (grap[i].size() % 2 == 0)
            even++;
        else
            odd++;
    }
    cout << endl;
    DFS(1);
    if (s.size() != n)
        cout << "Non-Eulerian" << endl;
    else {
        if (even == n)
            cout << "Eulerian" << endl;
        else if (odd == 2)
            cout << "Semi-Eulerian" << endl;
        else
            cout << "Non-Eulerian" << endl;
    }
    return 0;
}