maxflow

DescriptionEvery time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. this means that the clover is covered by water for awhile and takes quite a long time to regrow. thus, Farmer John has built a set of drainage

are printed from the beginning of a separate line.Sample Input2 1 1 2 (0,1) (1,0) ten (0) (1) 207 2 3 (0,0) 1 (0,1) 2 (0,2) 5 (1,0) 1 (8) (2,3) 1 (2,4) 7 (3,5) 2 (3,6) 5 (4,2) 7 (4,3 ) 5 (4,5) 1 (6,0) 5 (0) 5 (1) 2 (3) 2 (4) 1 (5) 4Sample Output156HintThe sample input contains the data sets. The first data set encodes a network with 2 nodes, power station 0 with Pmax (0) =15 and Consumer 1 with Cmax (1) =20, and 2 p Ower Transport lines with Lmax (0,1) =20 and Lmax (1,0) =10. Th

Ek (edmondskarp) algorithm: this algorithm is improved by the Ford-Fulkerson Algorithm. The Ford-Fulkerson Algorithm is constantly used to search for an augmented path, and then judge the minimum traffic of a single path, next, we add a stream to this path. What is different from the FF algorithm is that we use an array to record the minimum traffic of the extended path after extensive search, and then add the stream based on the parent array, time Complexity: O (VE ^ 2) Typedef struct {int flo

EdgeAdd (Inp[i].x,inp[i].y + N,0); Add (inp[i].x+n,inp[i].y, INF); Add (inp[i].y,inp[i].x+ N,0); } intx, y; for(inti =1; I ) if(I! = s i! =t) Add (i,i+n,1), add (I+n,i,0);//set the Benquan between the split points to 1 intTMP =0, Maxflow =0; S+=N; while(BFS ())//maximum flow minimum cut while(TMP = Dinic (s,inf)) Maxflow + =tmp; Ans=min (ans,

Finally, I have finished writing the template of the network stream. I have to paste a few files and use the forward star to save and save the trouble of writing the linked list, the code is definitely the most streamlined in the code you can find // Ek# Include # Include Using namespace STD;# Include # Define maxn300# Define maxflow 2000000000Int N, S, T, M, flow [maxn + 1] [maxn + 1], map [maxn + 1] [maxn + 1], pre [maxn + 1];Void Init (){Int I, A,

http://poj.org/problem?id=1966Title Description:In cable TV networks, repeaters are connected in two directions. If there is at least one route between any two repeaters in the network, the repeater network is called connected, or the repeater network is not connected. An empty network, and a network with only one repeater, are considered to be connected. The safety factor F of a network with n repeaters is defined as:(1) F is n, the remaining network is still connected, regardless of the number

);return m-2;}int augment () {int x=t,a=inf;While (x!=s) {Edge e=edges[p[x]];a=min (a,e.cap-e.flow);X=edges[p[x]].from;}x=t;While (x!=s) {Edges[p[x]].flow+=a;Edges[p[x]^1].flow-=a;X=edges[p[x]].from;}return A;}int maxflow (int s,int t) {int flow=0;this->s=s; this->t=t;CLR (d,0); CLR (num,0); CLR (cur,0); CLR (p,0) ;Rep (I,edges.size ()) edges[i].flow=0;Rep (i,n) num[d[i]]++;int x=s;While (d[s]if (x==t) {flow+=augment (); x=s;}int ok=0;Rep (I,cur[x],g[

Recently when looking at MATLAB code, because I am using the 64 system, and code in the Mex file is compiled on the 32-bit system, so need to re-compile maxflowmex.cpp, but the direct Mex Maxflowmex.cpp, the following error occurred:Maxflowmex.obj:error LNK2019: unresolved external symbol "Public: __cdecl graphThe problem is due to the inability to compile multiple CPP files. The maxflowmex.cpp contains the head and shoulders as # include "Maxflow-v3.

graph can no longer increase traffic.If there is an augmented path, I will increase the maximum traffic while modifying the Edge CF (U,V) on the augmented path, and then finding the augmented path again.The whole process is probably:while (Findaugmentpath ())//Determine if there is an augmented path Maxflow = maxflow + Delta//MAX Stream increase modifygraph ()//Modify the augmentation path end whileLittle

[ from]=0; Avis[ from]=1; the while(!q.empty ()) + { - intu=Q.front (); Q.pop (); $vis[u]=0; $ for(intI=head[u]; i!=-1; i=edge[i].next) - { - intv=edge[i].v; the if(edge[i].flow>0 dis[v]>dis[u]+edge[i].cost) - {Wuyidis[v]=dis[u]+Edge[i].cost; thepre[v]=u; -pid[v]=i; Wu if(!Vis[v]) - { Aboutvis[v]=1; $ Q.push (v); - } - } - } A } + returnDis[to]; the }

No sap, and the adjacent matrix is too large. # Include # Include # Include # Include # Define maxn375# Define INF 0x3f3f3fUsing namespace STD;Int CAP [maxn + 1] [maxn + 1], flow [maxn + 1] [maxn + 1], N;Int P [maxn + 1], C [maxn + 1];Int seq [25] [15], maxflow;Inline void Init (){Maxflow = 0;Memset (Cap, 0, sizeof (CAP ));Memset (flow, 0, sizeof (flow ));}Inline int min (int A, int B){Return a }Void BFS ()

, i.e. U-->v if(E[p].flow !) e[p+2].flow) {U.push_back (e[p ^1].V-(I-1) * n); V.push_back (E[P].V-I * n); }//Reverse flow, positive no flow, that is, v-->u, so that the two sides have to exclude the flow of the situation, the equivalent of two points of exchange, that is, both ships stay in situ Else if(! E[p].flow E[p +2].flow) {u.push_back (e[(p +2) ^1].V-(I-1) * n); V.push_back (E[p +2].v-i * N); } }printf("%d", U.size ()); for(intp =0; P for(intj =1;

edges are full stream (that is, maxflow== all the du[]>0). Topic Link: sgu194 reactor cooling #include 2, there are Yuanhui with the upper and lower bounds of the maximum flow problem thinking: Add a source Point St, meeting point Sd,st to day I even one of the upper bound of Di lower bound to 0, each goddess to the meeting point of the lower bound to the GI upper bound for the OO side, for each day, the day to the first girl even a [Li,ri] sid

[I] = INF;Q. push (s );While (! Q. empty ()){Int now = q. front ();Q. pop ();If (now = n) break;For (int I = 1; I If (! Vis [I] g [now] [I]> 0 ){Vis [I] = true;Flow [I] = min (flow [now], g [now] [I]);Pre [I] = now;Q. push (I );}}}If (! Vis [e] | e = 1) // The complete augmenting or Source Vertex coincidence cannot be found.Return-1;ElseReturn flow [e];} Int EK (int s, int e ){Int temp, d, res, maxflow;Maxflow