PAT甲级 1126 - Eulerian Path

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根据题目所给的条件进行判断即可。

1. It has been proven that connected graphs with all vertices of even degree have an Eulerian circuit, and such graphs are called Eulerian. 在图联通的前提下，节点的入度和出度均为偶数。

2. If there are exactly two vertices of odd degree, all Eulerian paths start at one of them and end at the other. 在图联通的前提下，节点的入度和出度exactly两个奇数。

3. A graph that has an Eulerian path but not an Eulerian circuit is called semi-Eulerian 其他的就是费Eulerian path。

#include<bits/stdc++.h>
using namespace std;
const int maxn=500+3;
vector<int> edge[maxn];
int degree[maxn];
bool visited[maxn];
void dfs(int cur){
  for(vector<int>::iterator it = edge[cur].begin();it!=edge[cur].end();++it){
    if(!visited[*it]){
      visited[*it]=true;
      dfs(*it);
    }
  }
}
bool isConnected(int n){
  dfs(1);
  visited[1]=true;
  for(int i=1;i<=n;++i){
    if(!visited[i])return false;
  }
  return true;
}

int main(){
  int n,m;
  scanf("%d %d",&n,&m);
  int u,v;
  for(int i=0;i<m;++i){
    scanf("%d %d",&u,&v);
    edge[u].push_back(v),edge[v].push_back(u);
    ++degree[v],++degree[u];
  }
  int cntOdd=0;
  bool flag = isConnected(n);
  for(int i=1;i<=n;++i){
    if(i==1)printf("%d",degree[i]);
    else printf(" %d",degree[i]);
    cntOdd+=(degree[i]%2);
  }
  printf("\n");
  if(cntOdd==0 && flag){
    printf("Eulerian");
  }else if(cntOdd==2 && flag){
    printf("Semi-Eulerian");
  }else{
    printf("Non-Eulerian");
  }
  return 0;
}