hdu3849-By Recognizing These Guys, We Find Social Networks Useful：双连通分量

浏览数：48 / 时间：2015年06月09日

By Recognizing These Guys, We Find Social Networks Useful

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 2354 Accepted Submission(s): 613

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

Source

2011 Invitational Contest Host by BUPT

题意：有n个人名和m条边（用人名来表示），求出这个图中的所有桥（以人名表示边来输出）。

算法：用map来hash，边(a,b)的hash值为a*10000+b，然后求桥，最后按输入顺序遍历一遍所有边，如果为桥的话就输出。

此题有一个坑就是当图不连通的时候直接输出0就可以了。

  1 #include <iostream>
  2 #include <stdio.h>
  3 #include <map>
  4 #include <memory.h>
  5 #include <vector>
  6 using namespace std;
  7 
  8 
  9 const int maxn = 10000 + 10;
 10 int low[maxn],pre[maxn],dfs_clock=0;
 11 map<int,bool> isbridge;
 12 vector<int> G[maxn];
 13 int cnt_bridge;
 14 int father[maxn];
 15 
 16 int getid(int u,int v)
 17 {
 18     return u*10000+v;
 19 }
 20 
 21 int dfs(int u, int fa)
 22 {
 23     father[u]=fa;
 24     int lowu = pre[u] = ++dfs_clock;
 25     int child = 0;
 26     for(int i = 0; i < G[u].size(); i++)
 27     {
 28         int v = G[u][i];
 29         if(!pre[v])   // 没有访问过v
 30         {
 31             child++;
 32             int lowv = dfs(v, u);
 33             lowu = min(lowu, lowv); // 用后代的low函数更新自己
 34             if(lowv > pre[u]) // 判断边(u,v)是否为桥
 35             {
 36                 isbridge[getid(u,v)]=isbridge[getid(v,u)]=true;
 37                 cnt_bridge++;
 38             }
 39         }
 40         else if(pre[v] < pre[u] && v != fa)
 41         {
 42             lowu = min(lowu, pre[v]); // 用反向边更新自己
 43         }
 44     }
 45     return low[u]=lowu;
 46 }
 47 
 48 void init(int n)
 49 {
 50     isbridge.clear();
 51     memset(pre,0,sizeof pre);
 52     cnt_bridge=dfs_clock=0;
 53     for(int i=0; i<n; i++)
 54     {
 55         G[i].clear();
 56     }
 57 }
 58 
 59 
 60 bool vis[maxn];
 61 int cnt;
 62 int dfs_conn(int u)
 63 {
 64     vis[u]=true;
 65     cnt++;
 66     for(int i=0;i<G[u].size();i++)
 67     {
 68         int v=G[u][i];
 69         if(!vis[v])
 70             dfs_conn(v);
 71     }
 72 }
 73 
 74 bool isconn(int n)
 75 {
 76     memset(vis,false,sizeof vis);
 77     cnt=0;
 78     dfs_conn(0);
 79     return cnt==n;
 80 }
 81 
 82 
 83 int main()
 84 {
 85     #ifndef ONLINE_JUDGE
 86     freopen("in.txt","r",stdin);
 87     #endif
 88 
 89     int T;
 90     cin>>T;
 91     while(T--)
 92     {
 93         map<string,int> id;
 94         map<int,string> id2;
 95         vector<int> edges;
 96         int n,m;
 97         scanf("%d %d",&n,&m);
 98         init(n);
 99         int num=0;
100         for(int i=0;i<m;i++)
101         {
102 
103             char str1[20],str2[20];
104             scanf("%s %s",str1,str2);
105             int a,b;
106             if(id.count((string)str1)>0)
107             {
108                 a=id[(string)str1];
109             }
110             else
111             {
112                 a=id[(string)str1]=num++;
113             }
114 
115             if(id.count((string)str2)>0)
116             {
117                 b=id[(string)str2];
118             }
119             else
120             {
121                 b=id[(string)str2]=num++;
122             }
123 
124             id2[a]=(string)str1;
125             id2[b]=(string)str2;
126 
127             G[a].push_back(b);
128             G[b].push_back(a);
129             edges.push_back(getid(a,b));
130         }
131 
132         if(!isconn(n))
133         {
134             puts("0");
135             continue;
136         }
137 
138         dfs(0,-1);
139         cout<<cnt_bridge<<endl;
140         for(int i=0;i<edges.size();i++)
141         {
142             if(isbridge[edges[i]])
143             {
144                 printf("%s %s\n",id2[edges[i]/10000].c_str(),id2[edges[i]%10000].c_str());
145             }
146         }
147     }
148 
149     return 0;
150 }