Problem Description

Social
Network is popular these days.The Network helps us know about those
guys who we are following intensely and makes us keep up our pace with
the trend of modern times.
But how?
By what method can we know the
infomation we wanna?In some websites,maybe Renren,based on social
network,we mostly get the infomation by some relations with those
"popular leaders".It seems that they know every lately news and are
always online.They are alway publishing breaking news and by our
relations with them we are informed of "almost everything".
(Aha,"almost everything",what an impulsive society!)
Now,it's
time to know what our problem is.We want to know which are the key
relations make us related with other ones in the social network.
Well,what is the so-called key relation?
It
means if the relation is cancelled or does not exist anymore,we will
permanently lose the relations with some guys in the social
network.Apparently,we don't wanna lose relations with those guys.We must
know which are these key relations so that we can maintain these
relations better.
We will give you a relation description map and you should find the key relations in it.
We
all know that the relation bewteen two guys is mutual,because this
relation description map doesn't describe the relations in twitter or
google+.For example,in the situation of this problem,if I know you,you
know me,too.

 

Input

The input is a relation description map.
In the first line,an integer t,represents the number of cases(t <= 5).
In
the second line,an integer n,represents the number of guys(1 <= n
<= 10000) and an integer m,represents the number of relations between
those guys(0 <= m <= 100000).
From the second to the (m +
1)the line,in each line,there are two strings A and B(1 <=
length[a],length[b] <= 15,assuming that only lowercase letters
exist).
We guanrantee that in the relation description map,no one has
relations with himself(herself),and there won't be identical
relations(namely,if "aaa bbb" has already exists in one line,in the
following lines,there won't be any more "aaa bbb" or "bbb aaa").
We
won't guarantee that all these guys have relations with each other(no
matter directly or indirectly),so of course,maybe there are no key
relations in the relation description map.

 

Output

In the first line,output an integer n,represents the number of key relations in the relation description map.
From the second line to the (n + 1)th line,output these key relations according to the order and format of the input.

 

Sample Input

 

Sample Output

 

Source

2011 Invitational Contest Host by BUPT

 

 

 #include <iostream>

 #include <stdio.h>

 #include <map>

 #include <memory.h>

 #include <vector>

 using namespace std;

 const int maxn =  + ;

 int low[maxn],pre[maxn],dfs_clock=;

 map<int,bool> isbridge;

 vector<int> G[maxn];

 int cnt_bridge;

 int father[maxn];

 int getid(int u,int v)

 {

     return u*+v;

 }

 int dfs(int u, int fa)

 {

     father[u]=fa;

     int lowu = pre[u] = ++dfs_clock;

     int child = ;

     for(int i = ; i < G[u].size(); i++)

     {

         int v = G[u][i];

         if(!pre[v])   // 没有访问过v

         {

             child++;

             int lowv = dfs(v, u);

             lowu = min(lowu, lowv); // 用后代的low函数更新自己

             if(lowv > pre[u]) // 判断边(u,v)是否为桥

             {

                 isbridge[getid(u,v)]=isbridge[getid(v,u)]=true;

                 cnt_bridge++;

             }

         }

         else if(pre[v] < pre[u] && v != fa)

         {

             lowu = min(lowu, pre[v]); // 用反向边更新自己

         }

     }

     return low[u]=lowu;

 }

 void init(int n)

 {

     isbridge.clear();

     memset(pre,,sizeof pre);

     cnt_bridge=dfs_clock=;

     for(int i=; i<n; i++)

     {

         G[i].clear();

     }

 }

 bool vis[maxn];

 int cnt;

 int dfs_conn(int u)

 {

     vis[u]=true;

     cnt++;

     for(int i=;i<G[u].size();i++)

     {

         int v=G[u][i];

         if(!vis[v])

             dfs_conn(v);

     }

 }

 bool isconn(int n)

 {

     memset(vis,false,sizeof vis);

     cnt=;

     dfs_conn();

     return cnt==n;

 }

 int main()

 {

     #ifndef ONLINE_JUDGE

     freopen("in.txt","r",stdin);

     #endif

     int T;

     cin>>T;

     while(T--)

     {

         map<string,int> id;

         map<int,string> id2;

         vector<int> edges;

         int n,m;

         scanf("%d %d",&n,&m);

         init(n);

         int num=;

         for(int i=;i<m;i++)

         {

             char str1[],str2[];

             scanf("%s %s",str1,str2);

             int a,b;

             if(id.count((string)str1)>)

             {

                 a=id[(string)str1];

             }

             else

             {

                 a=id[(string)str1]=num++;

             }

             if(id.count((string)str2)>)

             {

                 b=id[(string)str2];

             }

             else

             {

                 b=id[(string)str2]=num++;

             }

             id2[a]=(string)str1;

             id2[b]=(string)str2;

             G[a].push_back(b);

             G[b].push_back(a);

             edges.push_back(getid(a,b));

         }

         if(!isconn(n))

         {

             puts("");

             continue;

         }

         dfs(,-);

         cout<<cnt_bridge<<endl;

         for(int i=;i<edges.size();i++)

         {

             if(isbridge[edges[i]])

             {

                 printf("%s %s\n",id2[edges[i]/].c_str(),id2[edges[i]%].c_str());

             }

         }

     }

     return ;

 }