对于u的子节点j，若有low[j] > dfn[u]，说明只要不走u-j这条路，从j的搜索树出发无法到达x或比x更早访问的节点

#include<bits/stdc++.h>

using namespace std;
const int N = 100010, M = N * 2;
int idx, h[N], ne[M], e[M], dfn[N], low[N], cnt;
bool bridge[M];
int n, m;
void add(int a, int b){
    e[idx] = b; ne[idx] = h[a]; h[a] = idx ++ ;
}

void tarjan(int u, int in_edge){
    dfn[u] = low[u] = ++ cnt;
    for(int i = h[u]; i != -1; i = ne[i]){
        int j = e[i];
        if(dfn[j] == 0){
            tarjan(j, i);

            low[u] = min(low[u], low[j]); // 回去时更新

            if(low[j] > dfn[u]){
                bridge[i] = bridge[i ^ 1] = true; // i 和 i ^ 1标记一对双向边
            }

        } else if(i != (in_edge ^ 1)){
            low[u] = min(low[u], dfn[j]); // 不是搜索树上的边
        }
    }
}

int main(){
    memset(h, -1, sizeof h);
    cin >> n >> m;
    idx = 2;
    while(m -- ){
        int a, b;
        cin >> a >> b;
        add(a, b);
        add(b, a);
    }
    tarjan(1 ,0);

    for(int i = 2; i < idx; i += 2){
        if(bridge[i]){
            printf("%d %d\n", e[i ^ 1], e[i]);
        }
    }

    return 0;
}


//9 11
//1 2
//1 5
//1 6
//2 3
//2 5
//3 4
//4 5
//6 7
//6 8
//6 9
//8 9