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.
有n个牧场，要从一个牧场走到另一个牧场，要求至少有两条独立的路，现在有m条路，求至少要新建多少路，使得任意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，但是可以经过同一个中间顶点，给出的图保证已经连通。
先缩点，新图就是一棵树，在树上添边，使得所有的点缩为一个双连通分量，
也就是（树上度数为1的点+1）/2
#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<math.h>
#include<iostream>
#include<vector>
#include<set>
#include<stack>
#include<map>
#include<queue>
typedef unsigned long long ull;
typedef long long ll;
using namespace std;
const int N=1e5+10;
const int prime=2333317;
const int INF=0x3f3f3f3f3f;
struct node
{
int v,next;
bool cut;
} eage[N*10];
int head[N];
int dfn[N],low[N],Belong[N];
int du[N];
int Stack[N];
int tot;
int Index,top;
int block;
int n,m;
void Add(int u,int v)
{
eage[top].v=v;
eage[top].cut=0;
eage[top].next=head[u];
head[u]=top++;
}
void Tarjan(int u,int pre)
{
dfn[u]=low[u]=++Index;
Stack[tot++]=u;
for(int i=head[u]; i!=-1; i=eage[i].next)
{
int v=eage[i].v;
if(v==pre)continue;
if(!dfn[v])
{
Tarjan(v,u);
low[u]=min(low[u],low[v]);
if(low[v]>dfn[u])
{
eage[i].cut=1;
eage[i^1].cut=1;
}
}
else if(low[u]>dfn[v])
low[u]=dfn[v];
}
int v;
if(low[u]==dfn[u])
{
block++;
do
{
v=Stack[--tot];
Belong[v]=block;
}
while(v!=u);
}
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
memset(head,-1,sizeof(head));
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
memset(Belong,0,sizeof(Belong));
memset(Stack,0,sizeof(Stack));
memset(du,0,sizeof(du));
Index=top=tot=0;
block=0;
while(m--)
{
int u,v;
scanf("%d%d",&u,&v);
Add(u,v);
Add(v,u);
}
Tarjan(1,0);
for(int u=1; u<=n; u++)
{
for(int i=head[u]; i!=-1; i=eage[i].next)
{
if(eage[i].cut)
du[Belong[u]]++;
}
}
int ans=0;
for(int i=1; i<=block; i++)
{
if(du[i]==1)ans++;
}
printf("%d\n",(ans+1)/2);
}
return 0;
}
来源：CSDN
作者：雲轩x
链接：https://blog.csdn.net/weixin_44184902/article/details/103835582