PTA 1126 Eulerian Path

 

无向连通图，输出每个顶点的度并判断Eulerian、Semi-Eulerian和Non-Eulerian这3种情况，我们直接记录每个点所连接的点，这样直接得到它的度，然后利用深度优先和visit数组来判断图是否连通，不连通直接是Non-Eulerian情况，连通的话再判断顶点的度是否都是偶数，是的话就是Eulerian情况，不是的话再判断它奇数度的个数是否为2个，也就是偶数比总顶点数少2个，是的话就是就是Semi-Eulerian情况了。

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
const int maxn = 1e9;
const int maxm = 1e5 + 5;
const ll inf = 2147483647;
using namespace std;
vector<int> v[505];
int visit[505];

void dfs(int x) {
    if (visit[x] == 0) {
        visit[x] = 1;
        for (int i = 0; i < v[x].size(); i++) {
            dfs(v[x][i]);
        }
    }
}

int main() {
    int n, m;
    cin >> n >> m;
    int a, b;
    for (int i = 0; i < m; i++) {
        cin >> a >> b;
        v[a].push_back(b);
        v[b].push_back(a);
    }
    int sum = 0;
    cout << v[1].size();
    if (v[1].size() % 2 == 0)sum++;
    for (int i = 2; i <= n; i++) {
        cout << ' ' << v[i].size();
        if (v[i].size() % 2 == 0)
            sum++;
    }
    cout << "\n";
    dfs(1);
    int f = 1;
    for (int i = 1; i <= n; i++) {
        if (visit[i] == 0)
            f = 0;
    }
    if (!f) {
        cout << "Non-Eulerian";
        return 0;
    }
    if (sum == n)
        cout << "Eulerian";
    else if (sum == n - 2)
        cout << "Semi-Eulerian";
    else cout << "Non-Eulerian";
    return 0;
}