POJ 3177 Redundant Paths（边双连通的构造）

Redundant Paths

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13717		Accepted: 5824

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.

题目链接：POJ 3177

给你可能有多个的无向图，求最少加几条边使得图成为一个双连通分量。

做法：tarjan求出桥并把桥边标记删除，再DFS出各个连通块，并标注上所属块id，然后统计缩点之后的每个点的度，显然叶子节点的度为1即只连着一条边，但由于是无向图，会重复算一次，因此实际上所有点的度会多一倍，记叶子节点的个数为$leaf$，则答案为$(leaf+1)/2$，为什么是这样呢？因为当叶子为偶数的时候，可以找到具有最远LCA的叶子，两两之间连一条边形成环，这样便融入了双连通分量中而且这样是最优的方案，奇数的话剩下的一个随意找一个叶子融合一下就行了，这题据说有重边，用id处理一下即可。

代码：

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <sstream>
#include <numeric>
#include <cstring>
#include <bitset>
#include <string>
#include <deque>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=5010;
const int M=10010;
struct edge
{
    int to,nxt;
    int id,flag;
};
edge E[M<<1];
int head[N],tot;
int dfn[N],low[N],st[N],ts,scc,top,belong[N],deg[N];
bitset<N> ins;

void init()
{
    CLR(head,-1);
    tot=0;
    CLR(dfn,0);
    CLR(low,0);
    ts=scc=top=0;
    CLR(belong,0);
    CLR(deg,0);
    ins.reset();
}
inline void add(int s,int t,int id)
{
    E[tot].to=t;
    E[tot].id=id;
    E[tot].flag=false;
    E[tot].nxt=head[s];
    head[s]=tot++;
}
void Tarjan(int u,int id)
{
    dfn[u]=low[u]=++ts;
    ins[u]=1;
    st[top++]=u;
    int i,v;
    for (i=head[u]; ~i; i=E[i].nxt)
    {
        if(E[i].id==id)
            continue;
        v=E[i].to;
        if(!dfn[v])
        {
            Tarjan(v,E[i].id);
            low[u]=min(low[u],low[v]);
            if(low[v]>dfn[u])
            {
                E[i].flag=true;
                E[i^1].flag=true;
            }
        }
        else if(ins[v])
            low[u]=min(low[u],dfn[v]);
    }
    if(low[u]==dfn[u])
    {
        ++scc;
        do
        {
            v=st[--top];
            ins[v]=0;
        }while (u!=v);
    }
}
void dfs(int u,int x)
{
    belong[u]=x;
    ins[u]=1;
    for (int i=head[u]; ~i; i=E[i].nxt)
    {
        int v=E[i].to;
        if(!ins[v]&&!E[i].flag)
            dfs(v,x);
    }
}
int main(void)
{
    int n,m,a,b,i,j;
    while (~scanf("%d%d",&n,&m))
    {
        init();
        for (i=0; i<m; ++i)
        {
            scanf("%d%d",&a,&b);
            add(a,b,i);
            add(b,a,i);
        }
        for (i=1; i<=n; ++i)
            if(!dfn[i])
                Tarjan(i,-1);
        ins.reset();
        int block=0;
        for (i=1; i<=n; ++i)
            if(!ins[i])
                dfs(i,++block);
        for (i=1; i<=n; ++i)
        {
            for (j=head[i]; ~j; j=E[j].nxt)
            {
                int v=E[j].to;
                if(belong[i]!=belong[v])
                {
                    ++deg[belong[i]];
                    ++deg[belong[v]];
                }
            }
        }
        int ans=0;
        for (i=1; i<=block; ++i)
            if(deg[i]==2)
                ++ans;
        printf("%d
",(ans+1)>>1);
    }
    return 0;
}