PAT A1126 Eulerian Path （25 分）——连通图，入度-白红宇

PAT A1126 Eulerian Path （25 分）——连通图，入度

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发布时间：2019-06-12

本文共 2386 字，大约阅读时间需要 7 分钟。

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 )

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 (

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:

7 125 71 21 32 32 43 45 27 66 34 56 45 6

Sample Output 1:

2 4 4 4 4 4 2Eulerian

Sample Input 2:

6 101 21 32 32 43 45 26 34 56 45 6

Sample Output 2:

2 4 4 4 3 3Semi-Eulerian

Sample Input 3:

5 81 22 55 44 11 33 23 45 3

Sample Output 3:

3 3 4 3 3Non-Eulerian

1 #include 
       
         2 #include 
        
          3 using namespace std; 4 const int maxn=510; 5 int g[maxn][maxn]; 6 int deg[maxn]; 7 int n,m; 8 int num=0; 9 bool vis[maxn]={
     false};10 void dfs(int st){11     if(vis[st]==false){12         vis[st]=true;13         num++;14         for(int i=1;i<=n;i++){15             if(vis[i]==false && g[st][i]==1){16                 dfs(i);17             }18         }19     }20 }21 int main(){22     scanf("%d %d",&n,&m);23     for(int i=0;i

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注意点：直接根据题目字面意思实现就好了。首先看是不是连通图，再看入度为奇数的有几个，没有就是Eulerian，有2个就是semi，其余为non

转载于:https://www.cnblogs.com/tccbj/p/10431474.html

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