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

1
4 4
saerdna aswmtjdsj
aswmtjdsj mabodx
mabodx biribiri
aswmtjdsj biribiri

Sample Output

1
saerdna aswmtjdsj

題意：給出一個 n 個點 m 條邊的無向圖，求這個圖中橋的個數，並按順序輸出橋

思路：由於圖可能不連通，因此首先需要判斷是否存在 dfn[i]=0，如果存在則說明圖不連通，直接輸出 0 即可

求橋可以直接套 Tarjan 演算法模版，難點在於按輸入順序求橋，因此不能在求橋的過程中輸出，必須在求完所有橋後按順序判斷每條邊是否是橋，因此可用 map 來實現一個字串到點編號的對映，則一個輸入邊即為兩個字串結點 Node 組成

對於邊 x-y 來說，只要滿足 low[x]>dfn[y] 或 low[y]>dnf[x] 則該邊一定是橋，因而先求完所有節點的 low 值，然後在退出 Tarjan 演算法後，重新掃描每一條邊的兩個節點 x、y 的 low值、dfn 值，再判斷該邊是否為橋

Source Program

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
#include<cstdlib>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<ctime>
#include<vector>
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define N 20001
#define MOD 16007
#define E 1e-6
#define LL long long
using namespace std;
struct Node{
    char str[20];
    bool operator <(const Node s)const{
        return strcmp(str,s.str)<0;
    }
};
struct Edge{
    Node x;
    Node y;
    bool flag;//判斷該邊是否為割邊
}edge[N];
map<Node,int> mp;//將 字串Node對映成結點編號
vector<int> G[N];
int n,m;
int dfn[N],low[N];
int block_cnt;
void Tarjan(int x,int father){
    low[x]=dfn[x]=++block_cnt;

    for(int i=0;i<G[x].size();i++){
        int y=G[x][i];
        if(y==father)
            continue;

        if(dfn[y]==0){
            Tarjan(y,x);
            low[x]=min(low[x],low[y]);
        }
        else
            low[x]=min(low[x],dfn[y]);
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        mp.clear();
        for(int i=0;i<n;i++)
            G[i].clear();
        block_cnt=0;
        memset(dfn,0,sizeof(dfn));

        int num=0;//記錄編號
        for(int i=0;i<m;i++){
            edge[i].flag=false;;
            scanf("%s%s",edge[i].x.str,edge[i].y.str);

            if(mp.find(edge[i].x)==mp.end())
                mp[edge[i].x]=++num;
            if(mp.find(edge[i].y)==mp.end())
                mp[edge[i].y]=++num;

            int x=mp[edge[i].x];
            int y=mp[edge[i].y];
            G[x].push_back(y);
            G[y].push_back(x);
        }

        Tarjan(1,-1);
        bool connection=true;//判斷是否連通
        for(int i=1;i<=n;i++){
            if(dfn[i]==0){
                connection=false;
                break;
            }
        }

        if(connection==false)
            printf("0\n");
        else{
            int res=0;
            for(int i=0;i<m;i++){
                int x=mp[edge[i].x];
                int y=mp[edge[i].y];
                if(low[x]>dfn[y] || low[y]>dfn[x]){
                     edge[i].flag=true;
                     res++;
                }
            }

            printf("%d\n",res);
            for(int i=0;i<m;i++)
                if(edge[i].flag)
                    printf("%s %s\n",edge[i].x.str,edge[i].y.str);
        }
    }
    return 0;
}