1126 Eulerian Path (25 point(s))

 1126 Eulerian Path (25 point(s))

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

欧拉路径、欧拉回路的判断。

思路：先判断是否是连通图（并查集或者DFS），再统计每个结点的度。

参考链接：https://blog.youkuaiyun.com/richenyunqi/article/details/80169214

#include<iostream>
#include<cstring>
using namespace std;
const int MAX = 507;
int cnt[MAX]={0};
int Tree[MAX];
int N,M;
int findRoot(int x){
	if(Tree[x]==-1) return x;
	else{
		int temp = findRoot(Tree[x]);
		Tree[x] = temp;
		return temp;
	}
}
int main(void){
	cin>>N>>M;int a,b;
	memset(Tree,-1,sizeof(Tree));
	while(M--){
		cin>>a>>b;
		cnt[a]++;
		cnt[b]++;
		int na = findRoot(a);
		int nb = findRoot(b);
		if(na!=nb){
			Tree[na]=nb;
		}//并查集，检查是否只有一个连通分支 
	}
	int k=0;
	for(int i=1;i<=N;i++){
		if(Tree[i]==-1) k++;
	}//计算连通分支数 
	for(int i=1;i<=N;i++){
		if(i==1) cout<<cnt[i];
		else cout<<" "<<cnt[i]; 	
	}
	cout<<endl;
	if(k==1){//如果连通 
		int odd=0;
		for(int i=1;i<=N;i++){
			if(cnt[i]%2!=0)  odd++;
		}
		if(odd==0) cout<<"Eulerian";
		else if(odd==2) cout<<"Semi-Eulerian";
		else cout<<"Non-Eulerian";
	}
	else cout<<"Non-Eulerian";
	
	return 0;
}