POJ 3352 无向图的双联通分量。

                                                                                                Road Construction

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 7373		Accepted: 3697

Description

It's almost summer time, and that means that it's almost summer construction
time! This year, the good people who are in charge of the roads on the tropical
island paradise of Remote Island would like to repair and upgrade the various
roads that lead between the various tourist attractions on the island.

The roads themselves are also rather interesting. Due to the strange customs
of the island, the roads are arranged so that they never meet at intersections,
but rather pass over or under each other using bridges and tunnels. In this way,
each road runs between two specific tourist attractions, so that the tourists do
not become irreparably lost.

Unfortunately, given the nature of the repairs and upgrades needed on each
road, when the construction company works on a particular road, it is unusable
in either direction. This could cause a problem if it becomes impossible to
travel between two tourist attractions, even if the construction company works
on only one road at any particular time.

So, the Road Department of Remote Island has decided to call upon your
consulting services to help remedy this problem. It has been decided that new
roads will have to be built between the various attractions in such a way that
in the final configuration, if any one road is undergoing construction, it would
still be possible to travel between any two tourist attractions using the
remaining roads. Your task is to find the minimum number of new roads
necessary.

Input

The first line of input will consist of positive integers n and
r, separated by a space, where 3 ≤ n ≤ 1000 is the number of
tourist attractions on the island, and 2 ≤ r ≤ 1000 is the number of
roads. The tourist attractions are conveniently labelled from 1 to n.
Each of the following r lines will consist of two integers, v and
w, separated by a space, indicating that a road exists between the
attractions labelled v and w. Note that you may travel in either
direction down each road, and any pair of tourist attractions will have at most
one road directly between them. Also, you are assured that in the current
configuration, it is possible to travel between any two tourist
attractions.

Output

One line, consisting of an integer, which gives the minimum number of roads
that we need to add.

Sample Input

Sample Input 1
10 12
1 2
1 3
1 4
2 5
2 6
5 6
3 7
3 8
7 8
4 9
4 10
9 10

Sample Input 2
3 3
1 2
2 3
1 3

Sample Output

Output for Sample Input 1
2

Output for Sample Input 2
0

Source

题意：最少加多少条边可以使图变成双连通图。

解题思路：tarjin缩点，然后找出叶子节点的个数leaf，答案为（leaf+1)/2。

            不过tarjin缩点写的貌似有bug，先mark一下。

代码：

 

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int head[2000],low[2000],dfn[2000],Stack[2000],instack[2000],indexx,top,scc,belong[2000],tol;
struct node
{
        int next,to;
}edge[1000000];
void add(int u,int v)
{
        edge[tol].to=v;
        edge[tol].next=head[u];
        head[u]=tol++;
}
void tarjin(int u,int pre)
{
         int i,v;
         low[u]=dfn[u]=++indexx;
         Stack[top++]=u;
         for(i=head[u];i!=-1;i=edge[i].next)
         {
                  v=edge[i].to;
                  if(v==pre)continue;
                  if(!dfn[v])
                  {
                           tarjin(v,u);
                           if(low[u]>low[v])low[u]=low[v];
                  }
                  else if(low[u]>dfn[v])low[u]=dfn[v];
         }
         if(low[u]==dfn[u])
         {
                  scc++;
                  do
                  {
                           v=Stack[--top];
                           belong[v]=scc;
                  }while(v!=u);
         }
}

int in[2000];
void solve(int n)
{
        memset(dfn,0,sizeof(dfn));
        memset(belong,0,sizeof(belong));
        scc=indexx=top=0;
        for(int i=1;i<=n;i++)if(!dfn[i])tarjin(i,i);
        //cout<<scc<<endl;
        int leaf=0;
        for(int i=1;i<=n;i++)
        {
                for(int j=head[i];j!=-1;j=edge[j].next)
                {
                        int u=edge[j].to;
                        if(belong[i]!=belong[u])
                        {
                                in[belong[u]]++;
                                in[belong[i]]++;
                        }
                }
        }
       // for(int i=1;i<=scc;i++)cout<<in[i]<<" ";cout<<endl;
       for(int i=1;i<=scc;i++)if(in[i]==2)leaf++;
       printf("%d\n",(leaf+1)/2);
}
int main()
{
        int i,j,k,m,n;
        while(~scanf("%d%d",&n,&m))
        {
                memset(head,-1,sizeof(head));tol=0;
                while(m--)
                {
                        scanf("%d%d",&i,&j);
                        add(i,j);
                        add(j,i);
                }
                solve(n);
        }
        return 0;
}