53 lines
1.4 KiB
C++
53 lines
1.4 KiB
C++
#include <bits/stdc++.h>
|
||
using namespace std;
|
||
const int N = 5e5 + 10, M = 4e6 + 10;
|
||
typedef pair<int, int> PII;
|
||
#define x first
|
||
#define y second
|
||
|
||
// 链式前向星
|
||
int e[M], h[N], idx, w[M], ne[M];
|
||
void add(int a, int b, int c = 0) {
|
||
e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
|
||
}
|
||
|
||
int dfn[N], low[N], ts, root;
|
||
set<PII> _set;
|
||
|
||
void tarjan(int u, int fa) {
|
||
dfn[u] = low[u] = ++ts;
|
||
|
||
// 对比有向图强连通分量的代码，这里没有入栈的操作，原因是本题只想求割边，并不是想缩点，
|
||
// 怀疑缩点时就需要加入stk,in_stk等操作了，待验证
|
||
for (int i = h[u]; ~i; i = ne[i]) {
|
||
int v = e[i];
|
||
if (v == fa) continue;
|
||
if (!dfn[v]) {
|
||
tarjan(v, u);
|
||
low[u] = min(low[u], low[v]);
|
||
// 记录割边,论证过的结论
|
||
if (low[v] > dfn[u]) _set.insert({u, v}); // SET天生会排序，按first由小到大
|
||
} else
|
||
low[u] = min(low[u], dfn[v]);
|
||
}
|
||
}
|
||
|
||
int n, m;
|
||
int main() {
|
||
#ifndef ONLINE_JUDGE
|
||
freopen("P1656.in", "r", stdin);
|
||
#endif
|
||
memset(h, -1, sizeof h);
|
||
scanf("%d %d", &n, &m);
|
||
while (m--) {
|
||
int a, b;
|
||
scanf("%d %d", &a, &b);
|
||
if (a != b) add(a, b), add(b, a);
|
||
}
|
||
|
||
for (root = 1; root <= n; root++)
|
||
if (!dfn[root]) tarjan(root, -1);
|
||
|
||
for (auto i : _set) printf("%d %d\n", i.x, i.y);
|
||
return 0;
|
||
} |