Codeforces Beta Round #89 (Div. 2) E. Bertown roads（Tarjan、边双连通分量）

题目链接：http://codeforces.com/problemset/problem/118/E

思路：首先要判断图是否是边双连通，这个Tarjan算法可以判断，若low[v] > dfn[u],则说明边（u，v)是桥，从而这个图不是边双连通，然后发现在判断的时候已经访问了所有的顶点，顺便加入就可以了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <map>
#define REP(i, a, b) for (int i = (a); i < (b); ++i)
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
using namespace std;

const int MAX_N = (300000 + 100);
int N, M, cnt, bcc_count;
int low[MAX_N], dfn[MAX_N], vis[MAX_N], mark[MAX_N];
stack<int > S;
vector<int > g[MAX_N];
map<pair<int, int>, int> mp;
vector<pair<int, int > > edge;

bool Tarjan(int u, int fa)
{
	int tag = 0;
	low[u] = dfn[u] = ++cnt;
	vis[u] = 1;
	S.push(u);
	REP(i, 0, (int)g[u].size()) {
		int v = g[u][i];
		if (v == fa && !tag) { tag = 1;  continue; }
		if (!dfn[v]) {
			if (!Tarjan(v, u)) return false;
			low[u] = min(low[u], low[v]);
			if (low[v] > dfn[u]) return false;
			else {
				pair<int, int >PPI = make_pair(u, v);
				edge.push_back(PPI);
				mark[mp[PPI]] = 1;
			}
		}
		else if (vis[v]) {
			low[u] = min(low[u], dfn[v]);
			pair<int, int> PPI = make_pair(u, v);
			if (!mark[mp[PPI]]) {
				mark[mp[PPI]] = 1;
				edge.push_back(PPI);
			}
		}
	}
	return true;
}

int main()
{
	cin >> N >> M;
	FOR(i, 1, M) {
		int u, v; cin >> u >> v;
		g[u].push_back(v);
		g[v].push_back(u);
		mp[make_pair(u, v)] = i;
		mp[make_pair(v, u)] = i;
	}
	cnt = bcc_count = 0;
	memset(vis, 0, sizeof(vis));
	memset(mark, 0, sizeof(mark));
	memset(dfn, 0, sizeof(dfn));
	if (!Tarjan(1, -1)) { puts("0"); return 0; }
	REP(i, 0, (int)edge.size()) {
		printf("%d %d\n", edge[i].first, edge[i].second);
	}
	return 0;
}