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 descri_ption 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.
边的双连通分量与有向图的连通分量类似
双连通分量多适用与缩点成树
这里有一个结论
就是向一个图中填边
最少填（叶子节点+1/2）条边就可以使原图成为双联通分量图
#include<cstdio>
#include<iostream>
#include<cstring>
#include<map>
#include<stack>
using namespace std;
map<int,int>mp;
stack<int>st;
int first[5005];
int from[20005];
int nxt[20005];
int to[20005];
int dfn[20005];
int low[20005];
int belong[5005];
int du[5005];
int flag=0;
int id=0;
int num=0;
int cnt=0;
int ans=0;
int number=0;
int n,m;
void tarjan(int x,int last)
{
st.push(x);
dfn[x]=low[x]=++id;
for(int i=first[x];i!=-1;i=nxt[i])
{
int temp=to[i];
if(temp==last) continue;
if(x==1) number++;
if(!dfn[temp])
{
tarjan(temp,x);
low[x]=min(low[x],low[temp]);
}
else
{
low[x]=min(low[x],dfn[temp]);
}
}
if(dfn[x]==low[x])
{
num++;
while(st.top()!=x)
{
belong[st.top()]=num;
st.pop();
}
belong[st.top()]=num;
st.pop();
}
return ;
}
void solve()
{
for(int i=1;i<=cnt;i++)
{
int temp1=from[i],temp2=to[i];
if(belong[temp1]!=belong[temp2])
{
du[belong[temp1]]++;
}
}
for(int i=1;i<=n;i++)
{
if(du[i]==1) ans++;
}
}
int main()
{
number=0;
memset(first,-1,sizeof(first));
memset(nxt,-1,sizeof(nxt));
cnt=0;
scanf("%d%d",&n,&m);
for(int i=1;i<=m;i++)//add
{
int temp1,temp2;
scanf("%d%d",&temp1,&temp2);
if(temp1>temp2) swap(temp1,temp2);
if(mp[temp1*5000+temp2]==1) continue;
mp[temp1*5000+temp2]=1;
cnt++;
from[cnt]=temp1;
nxt[cnt]=first[temp1];
to[cnt]=temp2;
first[temp1]=cnt;
cnt++;
from[cnt]=temp2;
nxt[cnt]=first[temp2];
to[cnt]=temp1;
first[temp2]=cnt;
}
tarjan(1,-1);
solve();
printf("%d
",(ans+1)/2);
}