【PAT甲级】1126 Eulerian Path

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🔑中文翻译：欧拉路径
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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
1
2
3
4
5
6
7
8
9
10
11
12
13

Sample Output 1:

2 4 4 4 4 4 2
Eulerian
1
2

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
1
2
3
4
5
6
7
8
9
10
11

Sample Output 2:

2 4 4 4 3 3
Semi-Eulerian
1
2

Sample Input 3:

5 8
1 2
2 5
5 4
4 1
1 3
3 2
3 4
5 3
1
2
3
4
5
6
7
8
9

Sample Output 3:

3 3 4 3 3
Non-Eulerian
1
2

题意

在图论中，欧拉路径是图中的一条路径，该路径满足恰好访问每个边一次。

而欧拉回路是一条在同一顶点处开始和结束的欧拉路径。

如果一个连通图的所有顶点的度数都为偶数，那么这个连通图具有欧拉回路，且这个图被称为欧拉图。

如果一个连通图中有两个顶点的度数为奇数，其他顶点的度数为偶数，那么所有欧拉路径都从其中一个度数为奇数的顶点开始，并在另一个度数为奇数的顶点结束。

具有欧拉路径但不具有欧拉回路的图被称为半欧拉图。

现在，给定一个无向图，请你判断它是欧拉图、半欧拉图还是非欧拉图。

思路

1. 输入每条边，并记录每条边的度数。
2. 通过 dfs 来计算一共经过了多少结点，从而判断该图是否为连通图。
3. 输出该图中所有结点的度数。
4. 判断该图是否为连通图，如果是连通图还要进一步计算度数为奇数的结点，然后再输出该图是什么图。

代码

#include
using namespace std;

const int N = 510;
int n, m;
int d[N];
bool g[N][N], st[N];

//判断是否为连通图
int dfs(int u)
{
    st[u] = true; //标记该点

    int res = 1;	//计算经过了多少结点
    for (int i = 1; i <= n; i++)   //遍历临近结点
        if (!st[i] && g[u][i]) //判断该边是否存在
            res += dfs(i);

    return res;
}

int main()
{
    cin >> n >> m;

    //输入每条边
    for (int i = 0; i < m; i++)
    {
        int a, b;
        cin >> a >> b;
        g[a][b] = g[b][a] = true;
        d[a]++, d[b]++;  //更新度数
    }

    int cnt = dfs(1); //判断是否为连通图

    //输出每个点的度数
    cout << d[1];
    for (int i = 2; i <= n; i++)   cout << " " << d[i];
    cout << endl;

    //判断是否为欧拉回路
    if (cnt == n)
    {
        //计算度数为奇数的结点数量
        int s = 0;
        for (int i = 1; i <= n; i++)
            if (d[i] % 2)
                s++;

        if (s == 0)    puts("Eulerian");
        else if (s == 2)   puts("Semi-Eulerian");
        else puts("Non-Eulerian");
    }
    else    puts("Non-Eulerian");

    return 0;
}
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