【PAT-A】1126. Eulerian Path (25)

测试点4 是N=1的情况
测试点3 是考虑不连通的情况
这里找连通图的思路是 在每一个连通区域内只有一个点，它的编号小于所有与它直接连通的点的编号

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;
}