H3C ER3100 路由器 V201R012和V101R004 哪个版本更好 请做具体说明，非常感谢

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单位使用的是H3C ER3100 路由器 V201R012版本，近期ARP检测，扫描WAN网段，出现大量MAC重复，致使网络拥堵网速较慢。能查到所有冲突都是一个物理地址所致。但是在局域网内查找，该物理地址无对应主机。详细如下：
ARP检测 扫描网段： WAN 地址范围： －
序号 IP地址 MAC地址 接口 状态
1 123.81.255.129 00:18:82:1F:A7:AB WAN MAC重复
2 123.81.255.130 00:18:82:1F:A7:AB WAN MAC重复
3 123.81.255.132 00:18:82:1F:A7:AB WAN MAC重复
4 123.81.255.133 00:18:82:1F:A7:AB WAN MAC重复
5 123.81.255.134 00:18:82:1F:A7:AB WAN MAC重复
6 123.81.255.135 00:18:82:1F:A7:AB WAN MAC重复
7 123.81.255.136 00:18:82:1F:A7:AB WAN MAC重复
8 123.81.255.137 00:18:82:1F:A7:AB WAN MAC重复
9 123.81.255.138 00:18:82:1F:A7:AB WAN MAC重复
10 123.81.255.139 00:18:82:1F:A7:AB WAN MAC重复
11 123.81.255.163 00:18:82:1F:A7:AB WAN MAC重复
12 123.81.255.166 00:18:82:1F:A7:AB WAN MAC重复
13 123.81.255.172 00:18:82:1F:A7:AB WAN MAC重复
第 1 页/共 1 页 共 89 条记录 其中 89 条异常记录。
请高手指点。谢谢！

参考技术A 最新的不一定最好的 最旧的决定是又BUG的。一般推荐你使用正数第二个版本。 参考技术B 呵呵， 个人觉得还是拨打H3C的客服咨询，

因为H3C客服是最多接触你这样问题的 参考技术C side is always endless work, while the craving for the baby, not to mention the weight loss and skin care and exercise time to time, to return to the workplace, you suddenly find that time becomes your enemy, even if busy bruised and battered, is still how catch is not enough! At this time, effective time management can save you.
mother of four career will answer

1. To return to the workplace, the original work is suitable for you?
modern society,Ed hardy board shorts, not because you are female and you preferential treatment if the child is too young, you need to spend a lot of time to spend, then you need to consider the original post is also suitable for you? If the pregnancy before overtime, then transferred to more suitable to consider the status of your job.
2. To the knowledge of where to find child care?
successful career women each have their own work skills, empathy,Sandalia de Mujer MBT, each qualified child care mothers also have their own exclusive secrets. Of course, few people take care of the child's skills will be born, you also need to learn and explore. If you do not want to rush in when, after her mother, you need to learn early. In the district where you can join her mother obsessed with her mother and other classics, or some parenting websites, parenting forum, request support. Many mothers are eager helping hand extended.
3. Too busy mom, who is going to help?
baby after the birth of my father's work, entertainment is still busy, in respect of serious consideration: the special period, if they are still to be so busy? On hand which can delay things? What can be canceled? In many cases, busy life as a kind of inertia, so that we gradually become blind, as my father, also needs to change. Mom could not have been alone,Mens Ed hardy shirts Long, otherwise, too much pressure is likely to cause anxiety and depression.
4. How best to use parent-child time?
young parents' lives more colorful, in fact, to accompany the rich life and children are not in conflict, because the city life, we all need to absorb all kinds of new things, while looking for the release of a variety of channels pressure, however, after the parents or to try to accompany children. Busy moms and children to get along there is a trick: quality and not quantity. Emphasis is placed on communication quality, threw himself into playing with the children, observe the child's response and feedback, so that children feel your sincerity and love. Topics related articles：

birkenstock klum Xiang carrying baby daughter shot

[CF1499G]Graph Coloring

Graph Coloring

题解

一个晚上又没了。。。

我们考虑什么情况下 ∑ v ∈ V ∣ r ( v ) − b ( v ) ∣ \\sum_{v\\in V}\\left|r(v)-b(v)\\right| ∑v∈V​∣r(v)−b(v)∣最小。
很明显，这个值不可能小于 ∑ v ∈ V ∣ d e g v % 2 ∣ \\sum_{v\\in V}|deg_{v}\\%2| ∑v∈V​∣degv​%2∣，因为度数为奇数的点 ∣ r ( v ) − b ( v ) ∣ |r(v)-b(v)| ∣r(v)−b(v)∣最少为 1 1 1，它不可能让 r ( v ) = b ( v ) r(v)=b(v) r(v)=b(v)。
我们考虑如何让这个值刚好达到 d e g deg deg为奇数的点的个数。
我们可以尝试在这个图上构造路径，在路径上，我们对边交替染色，这样路径上非端点的点不会产生贡献，而端点会产生 1 1 1的贡献。
而必定存在一种染色方法使得所有 d e g deg deg为奇数的点为一条路径的端点，而偶数的点不为任何路径的端点。
因为如果对于偶数的点，它为端点的路径数一定是偶数，我们可以将经过它的端点们互相合并，最后它就没有路径出发了。而为奇数的点无论怎么消都会剩余一条路径，产生 1 1 1的贡献。
这样的话就刚好可以使图的权值最小。

但题目会动态加边并询问，我们必须要动态维护路径及其 h a s h hash hash值。
由于一个点最多作为一条路径的端点，我们没必要理会经过它的路径，只考虑它作端点的路径。当我们加入边 ( u , v ) (u,v) (u,v)时，有 3 3 3种情况：

  • 点 u , v u,v u,v都有路径出发，那么我们要加一条边，再将两条路径合并起来。
  • 点 u u u或点 v v v有一个有路径出发，那我们相当于对一条路径进行增广。
  • 点 u u u与点 v v v都不作端点，那我们相当于新加入一条路径。

而对于路径我们要维护其上的蓝色边 h a s h hash hash和，红色边 h a s h hash hash和，它的起终点与它的起终边。
由于合并路径时涉及到路径颜色的问题，相邻两条边需异色，如果合并的两条路径颜色相同我们可以直接将其合并，如果不同我们还要先将不同颜色的边交换后再合并。
考虑到合并的方便，我们可以先调整一下经过 u u u的路径的方向与经过 v v v的路径的方向。
但对于 t y p e = 2 type=2 type=2的询问，我们可以考虑先对于每条边记录下来它旁边的两条边，这同样可以再合并与增广时维护，再从一条路径的头开始， b f s / d f s bfs/dfs bfs/dfs找出所有的红边。
为了减少输出，我们可以将路径上偶数的边赋为蓝边 t y p e = 2 type=2 type=2的询问不会超过 10 10 10次呀，那没事了。

虽然每个操作 1 1 1都是 O ( 1 ) O\\left(1\\right) O(1)，但常数巨大无比。
当然，时间复杂度还是 O ( n + m + q ) O\\left(n+m+q\\right) O(n+m+q)的。
记得这是交互题哟！！！

源码

贞难调

#include<bits/stdc++.h>
using namespace std;
#define MAXN 1000005
#define lowbit(x) (x&-x)
#define reg register
#define pb push_back
#define mkpr make_pair
typedef long long LL;
typedef unsigned long long uLL;
const LL INF=0x7f7f7f7f7f7f7f7f;
const int mo=998244353;
const int jzm=2333;
const int orG=3,invG=332748118;
const double Pi=acos(-1.0);
typedef pair<int,int> pii;
const double PI=acos(-1.0);
template<typename _T>
_T Fabs(_T x){return x<0?-x:x;}
template<typename _T>
void read(_T &x){
	_T f=1;x=0;char s=getchar();
	while(s>'9'||s<'0'){if(s=='-')f=-1;s=getchar();}
	while('0'<=s&&s<='9'){x=(x<<3)+(x<<1)+(s^48);s=getchar();}
	x*=f;
}
template<typename _T>
void print(_T x){if(x<0){x=(~x)+1;putchar('-');}if(x>9)print(x/10);putchar(x%10+'0');}
int gcd(int a,int b){return !b?a:gcd(b,a%b);}
int add(int x,int y){return x+y<mo?x+y:x+y-mo;}
int qkpow(int a,int s){int t=1;while(s){if(s&1)t=1ll*a*t%mo;a=1ll*a*a%mo;s>>=1;}return t;}
int n1,n2,m,cnt,bl[MAXN],ans,pow2[MAXN],idx,Q,ch[MAXN][2],tim,vis[MAXN],dis[MAXN];
struct path{int s,t,rhs,bhs,siz,fir,til;}p[MAXN]; 
vector<int>Ans;queue<int>q;
void work(int u,int v){
	cnt++;pow2[cnt]=add(pow2[cnt-1],pow2[cnt-1]);
	if(bl[u]&&bl[v]&&bl[u]!=bl[v]){
		ans=add(ans,mo-add(p[bl[u]].rhs,p[bl[v]].rhs));
		if(p[bl[u]].s==u){swap(p[bl[u]].s,p[bl[u]].t);swap(p[bl[u]].fir,p[bl[u]].til);if(p[bl[u]].siz+1&1)swap(p[bl[u]].rhs,p[bl[u]].bhs);}
		if(p[bl[v]].t==v){swap(p[bl[v]].s,p[bl[v]].t);swap(p[bl[v]].fir,p[bl[v]].til);if(p[bl[v]].siz+1&1)swap(p[bl[v]].rhs,p[bl[v]].bhs);}
		if(p[bl[u]].siz&1)p[bl[u]].rhs=add(p[bl[u]].rhs,add(pow2[cnt],p[bl[v]].rhs)),p[bl[u]].bhs=add(p[bl[u]].bhs,p[bl[v]].bhs);
		else p[bl[u]].bhs=add(p[bl[u]].bhs,add(pow2[cnt],p[bl[v]].rhs)),p[bl[u]].rhs=add(p[bl[u]].rhs,p[bl[v]].bhs);
		ch[cnt][0]=p[bl[u]].til;ch[cnt][1]=p[bl[v]].fir;ch[p[bl[u]].til][ch[p[bl[u]].til][0]?1:0]=ch[p[bl[v]].fir][ch[p[bl[v]].fir][0]?1:0]=cnt;
		p[bl[u]].til=p[bl[v]].til;p[bl[u]].t=p[bl[v]].t,p[bl[u]].siz+=p[bl[v]].siz+1;p[bl[v]].siz=0;bl[p[bl[v]].t]=bl[u];
		ans=add(ans,p[bl[u]].rhs);if(v!=p[bl[u]].t)bl[v]=0;if(u!=p[bl[u]].s)bl[u]=0;
	}
	else if(bl[u]||bl[v]){
		if(bl[v])swap(u,v);ans=add(ans,mo-p[bl[u]].rhs);
		if(p[bl[u]].s==u){swap(p[bl[u]].s,p[bl[u]].t);swap(p[bl[u]].fir,p[bl[u]].til);if(p[bl[u]].siz+1&1)swap(p[bl[u]].rhs,p[bl[u]].bhs);}
		if(p[bl[u]].siz&1)p[bl[u]].rhs=add(p[bl[u]].rhs,pow2[cnt]);
		else p[bl[u]].bhs=add(p[bl[u]].bhs,pow2[cnt]);
		ch[p[bl[u]].til][ch[p[bl[u]].til][0]?1:0]=cnt;ch[cnt][0]=p[bl[u]].til;p[bl[u]].til=cnt;
		p[bl[u]].t=v;p[bl[u]].siz++;ans=addH3C校园网双出口配置


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