题目大意：给出一张无向图，任意两点可相互到达，现在要求你将这张无向图变成有向图，且改变之后任意两点还是可相互到达的
解题思路：桥的话肯定是要保留双向的，，所以可以在dfs的时候把桥标记出来，顺便在dfs的时候标记一下使用的是哪条边即可
Edge{int from, to, next, flag;}E[M];int head[N], pre[N], lowlink[N];int tot, n, m, dfs_clock;void AddEdge(int u, int v) {E[tot].from = u;E[tot].to = v;E[tot].flag = 0;E[tot].next = head[u];head[u] = tot++;}void init() {memset(head, -1, sizeof(head));tot = 0;int u, v;for (int i = 0; i < m; i++) {scanf(“%d%d”, &u, &v);AddEdge(u, v);AddEdge(v, u);}}void dfs(int u, int fa) {pre[u] = lowlink[u] = ++dfs_clock;int v;for (int i = head[u]; i != -1; i = E[i].next) {v = E[i].to;if (E[i].flag) continue;E[i].flag = 1;E[i ^ 1].flag = -1;if (!pre[v]) {dfs(v, u);lowlink[u] = min(lowlink[u], lowlink[v]);if (lowlink[v] > pre[u])E[i ^ 1].flag = 1;}else if (v != fa)lowlink[u] = min(lowlink[u], pre[v]);}}int cas = 1;void solve() {memset(pre, 0, sizeof(pre));dfs_clock = 0;for (int i = 1; i <= n; i++)if (!pre[i])dfs(i, -1);printf(“%d\n\n”, cas++);int u, v;for (int i = 0; i < tot; i++) {u = E[i].from;v = E[i].to;if (E[i].flag == 1)printf(“%d %d\n”, u, v);}printf(“#\n”);}int main() {while (scanf(“%d%d”, &n, &m) != EOF && n + m) {init();solve();}return 0;}
分明是比谁记的都清楚，比谁都更加在意，
