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## [$POJ$ $2230$ $Watchcow$](http://poj.org/problem?id=2230)
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### 一、大致题意：
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有一个人每晚要检查牛场，牛场内有$m$条路，他担心会有遗漏，就每条路检查两次，且每次的方向不同，要求你打印他行走的路径（必须从$1$开始），打印一条即可。
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### 二、实现代码
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```cpp {.line-numbers}
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#include <algorithm>
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#include <cstdio>
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#include <cstring>
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using namespace std;
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const int N = 10005, M = 2 * 50005;
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int n, m;
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// 链式前向星
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int e[M], h[N], idx, w[M], ne[M];
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void add(int a, int b, int c = 0) {
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e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
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}
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int res[M], rl;
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// 4个节点组成的无向图，这里面有10条边（满的是12条边，缺少了1和3之间的两条边）
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void dfs(int u) {
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// 无向图求欧拉回路,见边删边,点可以重复走，最终从1出发回到1，需要走10条边，共11个点
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for (int i = h[u]; ~i; i = h[u]) {
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h[u] = ne[i];
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dfs(e[i]);
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}
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res[++rl] = u;
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// 打点,可以直接输出，少用内存，还好想
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// printf("%d\n", u);
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}
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int main() {
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#ifndef ONLINE_JUDGE
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freopen("POJ2230.in", "r", stdin);
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#endif
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scanf("%d%d", &n, &m);
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memset(h, -1, sizeof h);
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int a, b;
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for (int i = 1; i <= m; i++) {
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scanf("%d%d", &a, &b);
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add(a, b), add(b, a);
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}
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dfs(1); // 题目要求从1号点出发
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for (int i = rl; i; i--) printf("%d\n", res[i]);
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return 0;
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}
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```
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