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## [$NC106112$ $Street$ $Directions$](https://www.luogu.com.cn/problem/UVA610)
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### 一、题目描述
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现在有一个联通的无向图，我们要把整个图改造为有向图，在保证强连通的情况下使得双向边尽可能少。
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### 二、解题思路
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- ① 桥不能被改成单向边，因为会使得某一个方向断开
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- ② 除了桥之外的边会构成若干个双连通图，那么我们按照$dfs$ 的顺序把这里面的边改成单向边，那么这个双连通图就会变成一个强连通图。
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**总结**
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- 桥，输出两条边
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- 不是桥，输出一条边，成对变换的另一个不能输出需要标识
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### 三、$Code$
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```cpp {.line-numbers}
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#include <bits/stdc++.h>
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using namespace std;
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const int N = 1010, M = N << 2;
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int ts;
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int dfn[N], low[N];
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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 out[M]; // 是不是成对变换输出过
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void tarjan(int u, int fa) {
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dfn[u] = low[u] = ++ts;
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for (int i = h[u]; ~i; i = ne[i]) {
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int v = e[i];
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if (v == fa) continue;
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if (out[i]) continue; // 对边已经输出过，那么，这条反边不能输出。因为如果是割边的话，两条边在u->v时就都已经输出完了
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out[i] = out[i ^ 1] = 1; // 标识成对变换输出过
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printf("%d %d\n", u, v); // 输出u->v,同时，需要检查 v->u是桥的话，还输出v->u
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if (!dfn[v]) {
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tarjan(v, u);
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low[u] = min(low[u], low[v]);
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if (dfn[u] < low[v]) // 割边
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printf("%d %d\n", v, u); // 割边需要输出两条
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} else
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low[u] = min(low[u], dfn[v]);
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}
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}
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int main() {
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#ifndef ONLINE_JUDGE
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freopen("UVA610.in", "r", stdin);
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#endif
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int n, m, cas = 0;
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while (scanf("%d%d", &n, &m), n + m) {
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idx = ts = 0;
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memset(h, -1, sizeof h);
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memset(dfn, 0, sizeof(dfn));
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memset(low, 0, sizeof(low)); // Tips:有些人的代码，low也是可以不用清空的
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memset(out, 0, sizeof out);
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while (m--) {
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int a, b;
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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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printf("%d\n\n", ++cas);
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for (int i = 1; i <= n; i++)
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if (!dfn[i]) tarjan(i, -1);
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puts("#");
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}
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return 0;
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}
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```
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