题解 | #欧拉回路#

#include <iostream>
using namespace std;
typedef long long ll;

const int N = 1010;
ll p[N], d[N], h[N], g[N];

ll Find(ll k) {
	if (p[k] != k) p[k] = Find(p[k]);
	return p[k];
}

void Union(ll a, ll b) {
	a = Find(a), b = Find(b);
	if (h[a] < h[b]) p[a] = b;
	else if (h[b] < h[a]) p[b] = a;
	else p[b] = a, h[a]++;
}

int main() {
	// freopen("input.txt", "r", stdin);
	int n, m, a, b, t;
	while (cin >> n && n) {
		cin >> m;
		for (int i = 0; i <= n; i++) p[i] = i, d[i] = 0;
		for (int i = 1; i <= m; i++) {
			cin >> a >> b;
			d[a]++, d[b]++;
			Union(a, b);
			t = a;// 记录其中一个连通图的点
		}
		int res = 0;
		bool flag = true;
		for (int i = 1; i <= n; i++) {
			if (Find(i) != Find(t) && d[i] != 0) {// 存在多个连同分量
				flag = false;
				break;
			} else if (d[i] & 1) {// 存在度不为偶数的点
				flag = false;
				break;
			}
		}
		if (flag) puts("1");
		else puts("0");
	}
	return 0;
}