poj1515--Street Directions（边的双连通）

给一个无向图，要求变成强连通的有向图，需要保留哪些边。

边的双连通，对于桥保留两条边，其他的只保留一条边。求双连通的过程中记录保留边。

/*********************************************
Problem: 1515		User: G_lory
Memory: 232K		Time: 32MS
Language: C++		Result: Accepted
**********************************************/
#include <cstdio>
#include <cstring>
#include <iostream>
#define pk printf("KKK!\n");

using namespace std;

const int N = 1005;
const int M = N * N;


struct Edge {
    int from, to, next;
    int cut;
} edge[M];
int cnt_edge;
int head[N];
void add_edge(int u, int v)
{
    edge[cnt_edge].from = u;
    edge[cnt_edge].to = v;
    edge[cnt_edge].next = head[u];
    edge[cnt_edge].cut = 0;
    head[u] = cnt_edge++;
}

int dfn[N]; int idx;
int low[N];

int n, m;

void tarjan(int u, int pre)
{
    dfn[u] = low[u] = ++idx;
    for (int i = head[u]; i != -1; i = edge[i].next)
    {
        int v = edge[i].to;
        if (edge[i].cut) continue;
        edge[i].cut = 1;
        edge[i ^ 1].cut = -1;
        if (v == pre) continue;

        if (!dfn[v])
        {
            tarjan(v, u);
            low[u] = min(low[u], low[v]);
            if (dfn[u] < low[v])
                edge[i].cut = edge[i ^ 1].cut = 1;
        }
        else low[u] = min(low[u], dfn[v]);
    }
}

void init()
{
    idx = cnt_edge = 0;
    memset(dfn, 0, sizeof dfn);
    memset(head, -1, sizeof head);
}

void solve()
{
    for (int i = 0; i < cnt_edge; ++i)
    if (edge[i].cut == 1)
    printf("%d %d\n", edge[i].from, edge[i].to);
}

int main()
{
    //freopen("in.txt", "r", stdin);
    int cas = 0;
    while (~scanf("%d%d", &n, &m))
    {
        if (n == 0 && m == 0) break;
        printf("%d\n\n", ++cas);
        int u, v;
        init();
        for (int i = 0; i < m; ++i)
        {
            scanf("%d%d", &u, &v);
            add_edge(u, v);
            add_edge(v, u);
        }
        tarjan(1, -1);
        solve();
        printf("#\n");
    }
    return 0;
}