AcWing 366. 看牛

题目链接：​​传送门​​

要求每条边只被经过一次，而且是双向边
那直接对每条边记一个vis然后dfs即可

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

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;
}