PAT-1126 Eulerian Path

本文介绍了一种用于判定无向图是否为欧拉图、半欧拉图或非欧拉图的算法。通过深度优先搜索(DFS)来验证图的连通性,并统计每个顶点的度数,以此为基础进行判定。

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1126 Eulerian Path(25 分)

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 EulerianSemi-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

    判断是否是欧拉图/半欧拉图,dfs判断连通性,统计各端点的度。

#include<stdio.h>
#include<set>
#include<algorithm>
using namespace std;

bool g[505][505];
int degree[505];
bool visited[505]={0};
int n,m;
void dfs(int v){
	visited[v]=true;
	for(int i=1;i<=n;i++){
		if(!visited[i]&&g[v][i]==true)
			dfs(i);
	}
}
int main(){

	scanf("%d %d",&n,&m);
	fill(g[0],g[504]+505,false);
	fill(degree,degree+n,0);

	for(int i=0;i<m;i++){
		int a,b;
		scanf("%d%d",&a,&b);
		g[a][b]=true;
		g[b][a]=true;
		degree[a]++;
		degree[b]++;
	}

	dfs(1);
	bool connected=true;
	for(int i=1;i<=n;i++){
		if(!visited[i]){
			connected=false;	
			break;
		}
	}
	int odd=0;
	for(int i=1;i<=n;i++){
		if(degree[i]%2==1)
			odd++;
		if(i!=1)printf(" ");
		printf("%d",degree[i]);
	}
	printf("\n");
	if(connected&&odd==0)
		printf("Eulerian");
	else if(connected&&odd==2)
		printf("Semi-Eulerian");
	else
		printf("Non-Eulerian");

	return 0;
}

 

<think>根据用户问题,用户询问networkx模块是否包含eulerian_path函数。根据引用[1]中提到的内容,确实有nx.eulerian_path(G)这个函数。此外,引用[2]介绍了networkx是用于图结构的Python包,引用[3]和[4]提供了更多算法信息,但直接相关的是引用[1]中的函数列表。 因此,回答是肯定的,并应提供基本用法示例。同时,按照指令,需要生成相关问题。 回答结构: 1. 直接回答用户问题,确认存在该函数。 2. 简要说明欧拉路径的概念。 3. 给出一个简单示例代码。 4. 生成相关问题。 注意:所有数学表达式使用$...$格式,独立公式用$$...$$。由于这里没有复杂的数学公式,主要关注代码和文字描述。</think>是的,NetworkX 模块提供了 `eulerian_path()` 函数,用于计算图中的欧拉路径。以下是详细说明和示例: ### 1. 函数说明 `networkx.eulerian_path(G)` 用于在图中查找欧拉路径(遍历图中每条边恰好一次的路径)。其特点包括: - **输入参数**:图对象 `G`(需为无向图或有向图) - **返回值**:生成器对象,按顺序产生路径中的边(格式:`(u, v)`) - **前置检查**:建议先用 `networkx.has_eulerian_path(G)` 检查图中是否存在欧拉路径[^1] ### 2. 欧拉路径基础 欧拉路径需满足以下条件之一: - **无向图**:图中恰有 0 或 2 个顶点的度为奇数 - **有向图**:除起点(出度=入度+1)和终点(入度=出度+1)外,所有顶点入度=出度 欧拉路径的存在性可通过公式验证: 设图 $G$ 的顶点集为 $V$,边集为 $E$,则欧拉路径存在的充要条件是连通且满足度约束[^2]。 ### 3. 使用示例 ```python import networkx as nx # 创建图(无向图示例) G = nx.Graph() G.add_edges_from([(1,2), (2,3), (3,4), (4,1), (1,3)]) # 检查是否存在欧拉路径 if nx.has_eulerian_path(G): # 获取欧拉路径 path = list(nx.eulerian_path(G)) print("欧拉路径:", path) # 输出: [(1, 2), (2, 3), (3, 1), (1, 4), (4, 3)] else: print("无欧拉路径") ``` ### 4. 相关函数 - `nx.is_eulerian(G)`:检查图是否为欧拉图(存在欧拉回路) - `nx.eulerian_circuit(G)`:计算欧拉回路 - `nx.eulerize(G)`:通过添加边使图欧拉化[^1]
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