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題目鏈接：傳送門

要求每條邊只被經過一次，而且是雙向邊
那直接對每條邊記一個vis然後dfs即可

#include <bits/stdc++.h>
#define A 100010

using namespace std;
struct node {int next, to;}e[A];
int head[A], num;
void add(int fr, int to) {e[++num].next = head[fr]; e[num].to = to; head[fr] = num;}
int n, m, a, b; bool vis[A]; vector<int> v;
void dfs(int fr) {
	for (int i = head[fr]; i; i = e[i].next) {
		int ca = e[i].to;
		if (vis[i]) continue;
		vis[i] = 1;
		dfs(ca); v.push_back(ca);
	}
}

int main(int argc, char const *argv[]) {
	cin >> n >> m;
	for (int i = 1; i <= m; i++) scanf("%d%d", &a, &b), add(a, b), add(b, a);
	dfs(1); v.push_back(1);
	for (auto i : v) cout << i << endl;
}