UVA610 Street Directions

题目大意

给出一个无向连通图，让你改造成一个有向图，并保证强连通。

解题思路

先思考，将无向图改成有向图，需要顾及到一些割边，因为把割边删掉后，原图就不联通了，所以改造后的有向图需要完整的保留割边。

而对于不是割边的边，我们只需要保留一条边，可以在 tarjan 时，记录下来。

需要注意的：多测不清空，亲人两行泪。

AC CODE

#include<bits/stdc++.h>
using namespace std;

struct Fastio
{
	template <typename T>
	inline Fastio operator>>(T &x)
	{
		x = 0;
		char c = getchar();
		while (c < '0' || c > '9')
			c = getchar();
		while (c >= '0' && c <= '9')
			x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
		return *this;
	}
	inline Fastio &operator<<(const char *str)
	{
		int cur = 0;
		while (str[cur])
			putchar(str[cur++]);
		return *this;
	}
	template <typename T>
	inline Fastio &operator<<(T x)
	{
		if (x == 0)
		{
			putchar('0');
			return *this;
		}
		if (x < 0)
			putchar('-'), x = -x;
		static int sta[45];
		int top = 0;
		while (x)
			sta[++top] = x % 10, x /= 10;
		while (top)
			putchar(sta[top] + '0'), --top;
		return *this;
	}

} io;

#define _ 20005

int n, m, opt;

int tot = 1, head[_], from[_ << 1], to[_ << 1], nxt[_ << 1];

int dol[_];

int cnt_node, cntn, low[_], dfn[_], id[_], vis[_ << 1];
stack<int> s;

int u[_], v[_];

int js(int x)
{
	return (x % 2) ? x + 1 : x - 1;
}

void add(int u, int v)
{
	from[++tot] = u;
	to[tot] = v;
	nxt[tot] = head[u];
	head[u] = tot;
}

void tarjan(int u, int fa)
{
	low[u] = dfn[u] = ++cnt_node;
	s.push(u);
	for(int i = head[u]; i; i = nxt[i])
		if(vis[i] == -1)
		{
			if(to[i] == fa) continue;
			vis[i] = 1;
			vis[i ^ 1] = 0;
			if(!dfn[to[i]])
			{
				tarjan(to[i], u);
				low[u] = min(low[u], low[to[i]]);
				if(low[to[i]] > dfn[u]) vis[i ^ 1] = 1;
			}
			else low[u] = min(low[u], dfn[to[i]]);
		}
}

void init()
{
	tot = 1;
	cntn = 0;
	cnt_node = 0;
	memset(dfn, 0, sizeof dfn);
	memset(low, 0, sizeof low);
	memset(id, 0, sizeof id);
	while(!s.empty()) s.pop();
	memset(u, 0, sizeof u);
	memset(v, 0, sizeof v);
	memset(head, 0, sizeof head);
	memset(vis, -1, sizeof vis);
}

signed main()
{
	while(1)
	{
		io >> n >> m;
		if(!n && !m) break;
		init();
		
		for(int i = 1; i <= m; ++i)
		{
			io >> u[i] >> v[i];
			add(u[i], v[i]);
			add(v[i], u[i]);
		}
		io << ++opt << "\n" << "\n";
		for(int i = 1; i <= n; ++i)
			if(!dfn[i]) tarjan(i, 0);
		for(int i = 2; i <= tot; ++i)
  		{
			if(vis[i])
   			{
   				io << from[i] << " " << to[i] << "\n";
   			}
		}
		io << "#" << "\n";
	}
	return 0;
}