无向连通图至少增加多少边形成双联通

poj3177

Redundant Paths

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8558		Accepted: 3712

Description

In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the rest of the herd are forced to cross near the Tree of Rotten Apples. The cows are now tired of often being forced to take a particular path and want to build some new paths so that they will always have a choice of at least two separate routes between any pair of fields. They currently have at least one route between each pair of fields and want to have at least two. Of course, they can only travel on Official Paths when they move from one field to another. 

Given a description of the current set of R (F-1 <= R <= 10,000) paths that each connect exactly two different fields, determine the minimum number of new paths (each of which connects exactly two fields) that must be built so that there are at least two separate routes between any pair of fields. Routes are considered separate if they use none of the same paths, even if they visit the same intermediate field along the way. 

There might already be more than one paths between the same pair of fields, and you may also build a new path that connects the same fields as some other path.

Input

Line 1: Two space-separated integers: F and R 

Lines 2..R+1: Each line contains two space-separated integers which are the fields at the endpoints of some path.

Output

Line 1: A single integer that is the number of new paths that must be built.

Sample Input

7 7
1 2
2 3
3 4
2 5
4 5
5 6
5 7

Sample Output

2

Hint

Explanation of the sample: 

One visualization of the paths is: 

   1   2   3
   +---+---+  
       |   |
       |   |
 6 +---+---+ 4
      / 5
     / 
    / 
 7 +

Building new paths from 1 to 6 and from 4 to 7 satisfies the conditions. 

   1   2   3
   +---+---+  
   :   |   |
   :   |   |
 6 +---+---+ 4
      / 5  :
     /     :
    /      :
 7 + - - - - 

Check some of the routes: 
1 – 2: 1 –> 2 and 1 –> 6 –> 5 –> 2 
1 – 4: 1 –> 2 –> 3 –> 4 and 1 –> 6 –> 5 –> 4 
3 – 7: 3 –> 4 –> 7 and 3 –> 2 –> 5 –> 7 
Every pair of fields is, in fact, connected by two routes. 

It's possible that adding some other path will also solve the problem (like one from 6 to 7). Adding two paths, however, is the minimum.

题目大意：有F个牧场，1<=F<=5000，现在一个牧群经常需要从一个牧场迁移到另一个牧场。奶牛们已经厌烦老是走同一条路，所以有必要再新修几条路，这样它们从一个牧场迁移到另一个牧场时总是可以选择至少两条独立的路。现在F个牧场的任何两个牧场之间已经至少有一条路了，奶牛们需要至少有两条。
给定现有的R条直接连接两个牧场的路，F-1<=R<=10000，计算至少需要新修多少条直接连接两个牧场的路，使得任何两个牧场之间至少有两条独立的路。两条独立的路是指没有公共边的路，但可以经过同一个中间顶点

解题思路：给定一个无向连通图，判断至少需要加多少条边，使得任意两点之间至少有两条相互‘边独立’的道路，也就是说，至少加多少条边，使得这个图成为一个边双连通图。
首先我们已经有了一个连通图
那么我们将所有的双连通块看成一个点，这样可以得到一颗树，因为他们当中没有环，而又是一个连通的图，所以我们可以肯定的是，这样缩点之后可以得到一颗树，这样的树至少需要加多少条边就能构成一个双连通图呢，我们只需要将叶子节点连起来即可，所以我们最少要连的边=（叶子节点数+1)/2

利用low的性质在一个连通块中的low值是相同的
而求边连通分量，用tarjan就可以求出哪个点属于哪个双连通块中，然后再计算度为1的节点就可以了

程序：

#include"stdio.h"
#include"string.h"
#include"iostream"
#define M 10009
#define inf 99999999
using namespace std;
struct st
{
    int u,v,w,next;
}edge[50009];
int head[M],dfn[M],low[M],use[M],bridge[M],num,t,indx,suo[M],used[M];
int min(int a,int b)
{
    return a<b?a:b;
}
void init()
{
    t=0;
    memset(head,-1,sizeof(head));
}
void add(int u,int v)
{
    edge[t].u=u;
    edge[t].v=v;
    edge[t].next=head[u];
    head[u]=t++;
}
void tarjan(int u,int id)
{
    dfn[u]=low[u]=++indx;
    int i;
    for(i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(i==(id^1))continue;
        if(!dfn[v])
        {
            tarjan(v,i);
            low[u]=min(low[u],low[v]);
            if(low[v]>dfn[u])
            {
                bridge[num++]=i;
            }
        }
        else
        low[u]=min(low[u],dfn[v]);
    }
}
void solve(int n)
{
    num=indx=0;
    memset(dfn,0,sizeof(dfn));
    for(int i=1;i<=n;i++)
        if(!dfn[i])
        tarjan(i,-1);
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=-1)
    {
        init();
        while(m--)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            int flag=0;
            for(int i=head[a];i!=-1;i=edge[i].next)//去重边
            {
                int v=edge[i].v;
                if(b==v)
                    flag++;
            }
            if(flag)
                continue;
            add(a,b);
            add(b,a);
        }
        solve(n);
        memset(use,0,sizeof(use));
        int k=0;
        memset(suo,0,sizeof(suo));
        memset(used,0,sizeof(used));
        for(int i=1;i<=n;i++)//把原图进行缩点，形成一颗树节点为k个
        {
            if(!used[low[i]])
            {
                used[low[i]]++;
                suo[k++]=low[i];
            }
        }
        for(int i=0;i<t;i+=2)//求度数
        {
            int u=edge[i].u;
            int v=edge[i].v;
            if(low[u]!=low[v])
            {
                use[low[u]]++;
                use[low[v]]++;
            }
        }
        int ans=0;
        for(int i=0;i<k;i++)
        {

            if(use[suo[i]]==1)//统计入度为一的点的个数
                ans++;
        }
        printf("%d
",(ans+1)/2);
    }
    return 0;
}