【PAT-A】1126. Eulerian Path (25)

Code

// @author Birdy 2018.3.5
/*
1126
Point 3 连通性
Point 4 N = 1
*/
#include<iostream>
#include<algorithm>

using namespace std;

int main()
{
int N, M;
int num[502] = { 0 };
int root[502];
cin >> N >> M;
int H1, H2;
for (int i = 0; i < N; i++)
{
root[i] = i;
}
for (int i = 0; i < M; i++)
{
cin >> H1 >> H2;
num[H1 - 1]++;
num[H2 - 1]++;
root[H1 - 1] = min(root[H1 - 1], H2 - 1);
root[H2 - 1] = min(root[H2 - 1], H1 - 1);
}

int flag = 0;

for (int i = 0; i < N; i++)
{
if (0 == i)
cout << num[i];
else
{
cout << ' ' << num[i];
if (i == root[i])
flag = 3;
}

if (num[i] & 1 == 1)
flag++;
else if (0 == num[i])
flag = 3;

}

cout << endl;

if (flag == 0 || 1 == N)
cout << "Eulerian" << endl;
else if (flag == 2)
cout << "Semi-Eulerian" << endl;
else
cout << "Non-Eulerian" << endl;
system("pause");
return 0;
}