ccnp route questions

CCNP Study Notes 4-route part-route control◆ Passive-interface-can be used in rip ospf. In rip, there are two network commands in the Process of ospf, which only receive and not send packets. In the process of ospf, there are two network commands: 1, advertise the route of the network address segment. The mask length i

0101.0100.1.7.0------100.1.0000 0111.00000 0110---6Route-map Lo1 Permit 10Match IP Add 1Set Metric 5Route-map Lo1 per 20Match IP Add 2Set Metric 7Router RIPRedistribute Connect Route-map Lo1R3:sho IP ro5, no longer R2 do any configuration, so that R2 see these 8 routes, but you can see R1 R3 Lo0.R1:router RIPPassive-interface DefaultNeighbor 1.1.123.3The above implementation does not see 8 loopback1R3:int f0/0No IP Split-horizonAccess-list deny 100.1

-redistributionR3:Router OSPF 1Area 2 Nssa no-redistribution4 Special Areas, only Nssa does not automatically generate a default route, the title requires that the default next hop is R3, soR3:Router OSPF 1Area 2 Nssa Default-information-originateR1:sho IP roMetric values can be modified by area 2 Nssa default-information-originate metric 20, but not allowed in the title, so,R3:rouer OSPF 1Area 2 Default-cost 20R1; Sho ip ro os7, increase the lo3=30.1

+=addflow[t]*Dis[t]; intnow=T; while(now!=s) {Edg[pre[now]].flow-=Addflow[t]; Edg[pre[now]^1].flow+=Addflow[t]; now=edg[pre[now]].u; } return 1; }voidMCMF (intSintTLong Longflow,Long LongCost ) { while(SPFA (S,t,flow,cost));}voidinit () {Now_edge=0; memset (Head,-1,sizeof(head));} Mapstring,int>Map1;mapint,string>Map2;vectorint>ans;intN;voidDfsintx) { if(x==n)return; intu=x*2; for(inti=head[u];i!=-1; i=Edg[i].next) { if(edg[i^1].flowedg[i].v!=u-1) {Edg[i^1].flow--; Ans.push_ba

[maxn*MAXN];98 voidDfsintXBOOLQQ) About { - if(x==n)return;101 if(xN)102 {103 if(QQ) coutEndl;104DFS (x+n,qq); the if(!QQ) coutEndl;106 return;107 }108 for(intI=first[x];i;i=t[i].next)if(t[i].f==0!Vis[i])109 { thevis[i]=1;111 DFS (T[I].Y,QQ); the return;113 } the } the the voidInit ()117 {118scanf"%d%d",n,m);119 for(intI=1; i) - {121Cin>>S[i];122m[s[i]]=i;123 }124 for(intI=1; i) the {126 intX,y,tt;127Cin>

; - } - voidEX_GCD (ll A, ll B, ll x, LL y) the { - if(b)Wuyi { theEX_GCD (b, a%b, x, y); -LL tmp=x; Wux=y; -Y=tmp-(A/b) *y; About } $ Else - { -x=1, y=0; - return ; A } + } the ll C (ll N, ll M) - { $ if(N==m | | m==0) the return 1; the if(m==1|| m==n-1) the returnN; the //LL tmp=1, Ans=1; - //For (LL i=min (n, m); i>=1;i--) in // { the //tmp= (tmp*i)%mod; the //ans= (ans* (n+1-i))%mod; About // } the LL x, y; theEX_GCD (jc[m]*jc[n-m