network exam cost

DescriptionGiven a forward graph, each edge has a capacity of C and an expansion fee of W. The expansion fee here refers to the cost of expanding the capacity by 1.Please:1, in the case of no expansion, the maximum flow of 1 to N;2, the maximum flow of 1 to N to increase the minimum required for the expansion of K cost.InputThe first line contains three integer n,m,k, which represents the number of points, sides, and the amount of traffic required to

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EDMONDS-KARP: Multiple calls to BFS from S to T can be attempted to reduce the number of calls.  is the use of a less costly and efficient augmented approach.  consider: In an augmented process, look for multiple augmented paths. The implementation of dfs Dinic algorithm has the following steps: 1: Initialize capacity network and network Flow 2: constructs the residual

Reference:maximum stream + expense flow. first ask the maximum flow can be. The second question is "minimum cost maximum flow". by test instructions, this question can be translated into the "residual network" on the previous question, expanding the capacity of some edges so that the maximum flow from the new graph is K. so easy: for the side of the remaining traffic, the

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The first question directly runs the maximum flow can. The plan is built according to the cost flow, the cost is 0.For the second question, in the first question Dinic the remainder of the network to build, the original image of each side (I,J), built (I,J,INF,CIJ), that the cost of C can be augmented by the road. Then

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(Const intS) { inti,t,temp; Queueint>Q; memset (Dis,0x3f,sizeof(Dis)); Dis[s]=0; Visited[s]=true; Q.push (S); while(!Q.empty ()) {T=q.front (), Q.pop (); visited[t]=false; for(I=e.start (t); i;i=E[i].next) {Temp=e[i].to; if(E[I].W dis[t]+e[i].cDis[temp]) {Dis[temp]=dis[t]+e[i].c; if(!Visited[temp]) {Visited[temp]=true; Q.push (temp); } } } } returndis[ttt]!=0x3f3f3f3f;}intDfs (Const intSConst intBK) { if(S==TTT)returnBK; Visited[s]=true; intres

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Latest Network Administrator Exam outline I. Examination INSTRUCTIONS 1. Examination requirements (1) familiar with the basic knowledge of computer system; (2) familiar with the basic knowledge of data communication; (3) familiar with the architecture of computer network, understand the basic knowledge of TCP/IP protocol; (4) familiar with the common comp

Network Engineer Examination is the National computer technology and software level examination of an intermediate qualification examination, through the examination of qualified personnel in accordance with the requirements of the application Department of network system planning, design and installation of network equipment and hardware debugging work, can carr

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from:http://richardxx.yo2.cn/articles/Network maximum flow and minimum cost flow. html This period of time reviewed the network flow model, feeling than before the understanding has made great progress, although I know this thing is difficult to build on the model, and its algorithm itself is not very difficult, but I decided to say some of my understanding, afte

protocol (110 or less, as long as it is the same number on the line)R2 (config-router) #router-id 10.2.2.2//Assign IP to the route, similar to the identity cardR2 (config-router) #network 10.2.2.0 0.0.0.255 Area 0//Declare LO1, partitionR2 (config-router) #network 172.16.23.0 0.0.0.255 Area 0//Declare f1/0, partitionR2#show Run//View all ConfigurationsR3R45. Pick the right device, do RIP and OSPF redistrib