network graph software

].f>0) - { A inty=t[i].y; + if(dis[y]==dis[x]+1) the { - intA=ffind (Y,mymin (flow-now,t[i].f)); $t[i].f-=A; thet[t[i].o].f+=A; thenow+=A; the } the if(Now==flow) Break; - } in if(now==0) dis[x]=-1; the returnNow ; the } About the voidoutput () the { the for(intI=1; i2) +printf"%d->%d%d\n", T[I].X,T[I].Y,T[I].F); - } the Bayi intMax_flow () the { the intans=0; - while(BFS ()) - { theans+=Ffind (st,inf); the

imagine a pipe from the cross, there is water outflow from each column, Flow into each column. Since the problem itself is guaranteed to be solvable, it is a perfect match, and the water flowing out of the line will eventually be fully remitted to the into row.Then the intersection traffic of each row and column is the value of that point, because the range of the required values is 1~20, so we let the capacity of the row and column nodes be up to 19, and so on! Why is it 19? Because in the max

Topic linksTest Instructions: The number of cut points is given by a graph without direction .Ideas: very naked subject. You can do this directly by applying a template. Code:#include Copyright notice: This article Bo Master original articles, blogs, without consent may not be reproduced. Uva315-network (non-direction graph cut point)

Topic Link: Click to open the linkTest instructionsGiven a direction graph, ask:1) Select at least a few vertices. Ability to proceed from these vertices to reach all vertices2) At least how many edges to add. Talent makes it possible to reach all vertices from whatever vertexNumber of vertices After the strong connected component is obtained, the shrinkage point is calculated, and the degree of each point's penetration is computed.The answer to the f

graph theory algorithm-Network maximum flow template "Ek;dinic"ek TemplatesIncrease traffic by finding the minimum residue in the augmented residual network every timeconst int Inf=1e9;int n,m,s,t;struct node{int v,cap;}; vectorDinic templatesConstruct hierarchy + Block flow augmentationconst int Inf=1e9;int n,m;int s,t;int tot=1;struct node{int v,f,nxt;} E[10

communicate with each other by using buffers, producers can continue to produce the next data only by throwing data into the buffer, and consumers simply need to take the data from the buffer so that they do not block because of each other's processing speed. Support for non-uniform free and busyWhen the producer makes the data fast, the consumer is too late to deal with it, the unhandled data can be temporarily present in the buffer and slowly disposed of. Not because of consumer performan

Test instructions There is a C cow, each cow has the minimum and maximum value of the medicine it can accept, there is an L bottle of medicine, each bottle has a value of u and it can be used up to V cattle, to find out how many cows can be satisfied. Analysis: Network stream or binary graph multi-match or priority queue, this problem optimizes my dinic template, the original template will tle ... Code: PO

from all over the world who specialize in one type of aircraft, each of which has two drivers and requires a pilot and a co-pilot. For a variety of reasons, such as co-ordination, some pilots cannot fly on the same plane and ask how to match the driver to make the most of the flight., assuming that there are 10 drivers, the V1,V2,...,V10 represents up to 10 drivers, of which V1,V2,V3,V4,V5 is the pilot, and V6,V7,V8,V9,V10 is the co-pilot. If a pilot and a co-pilot can fly on the same plane, on

Codeforces 512C Fox And Dinner odd And even graph network flow, codeforces512c Question link: Click the open link Question: Given n people and their weights. Let people sit on any round table so that any two adjacent human rights values are prime numbers, and a table contains at least three people. Output The number of tables and the number of persons sitting at each table (output any specific scheme) Ideas

HDU 4888 Redraw Beautiful Drawings network flow graph, hduredraw Question: Given n, m, k The following n integers a [n] The following m integers B [n] Use numbers [0, k] to construct a matrix of n * m If a Unique solution exists, this matrix is output. If multiple solutions exist, Not Unique is output. If no solution exists, Impossible is output. Idea: network s

the degree of love to the minimum is how big?Problem Solving Ideas:Each shed does not live on a cow, so it is a multiple match. Match the time of the two-point enumeration of the degree of affection of the interval size, according to the interval size to enumerate the beginning and end of the interval, and then run multiple matches to determine whether it is legal. Note that the interval size is calculated. And, uh, when the data is read in, Maps[i][j] is not the i_th cow's favorite value for j

vertex appears only once (except for the starting point in a ring) so that the sum of the weights of all the edges in central China is minimized. (This problem does not indicate that there is no ring situation, directly according to the case of the ring can be done on the line)In order to become a ring, then the diagram is split, it becomes a one-way binary map, at this time a complete match is a connection strategy, as long as the guarantee that no edge is connected with themselves, can meet t

= head[u];k!=-1; k=Edge[k].next) { intv = edge[k].v,w=EDGE[K].W; if(level[v]==0w!=0) {Level[v]=level[u]+1; Q.push (v); } } } return-1;}intDfsintUintDesintincreaseroad) { if(U==des)returnIncreaseroad; intret=0; for(intk=head[u];k!=-1; k=Edge[k].next) { intv = edge[k].v,w=EDGE[K].W; if(level[v]==level[u]+1w!=0){ intmin = min (increaseroad-ret,w); W=DFS (v,des,min); EDGE[K].W-=W; Edge[k^1].w+=W; RET+=W; if(Ret==increaseroad)returnret; } }

will be followed by a line containing zero. OutputFor each test case, output a single integer on a line by itself-the number of seconds you and your friend need between the time he leaves the jail cell and the time both of you board the train. (assume that you do not need to wait for the train-they leave every second .) if there is no solution, print "back to jail ". Sample Input Sample output 211 2 999331 3 102 1 203 2 509121 2 101 3 101 4 102 5 103 5 104 5 105 7 106 7 107

Question link: Click the open link Question: Given the scores of n teams, output any feasible solution. There are only one and only one game for any two of the n teams, A and B. The competition results are divided into four types: 1. A + 3, B + 0 2. A + 0, B + 3 3. A + 2, B + 1 4. A + 1, B + 2 We found that in each result, two teams scored and always three. In four cases, they all split and form three. Therefore, we can combine any two teams into a single point. Connect N points to the source po

while(p!=S) { thef=min (f,cap[path[p]]); +p=to[path[p]^1]; -}p=T; $ while(p!=S) { $cap[path[p]]-=F; -cap[path[p]^1]+=F; -p=to[path[p]^1]; the } - returnF;Wuyi } the - intMCMF (intSintT) { Wu intv=0, D; - while((D=SPFA (s,t))! =INF) Aboutv+=d*(s,t); $ returnv; - } - }MCMF; - A ints,t; + intMain () { theFreopen ("starrace.in","R", stdin); -Freopen ("Starrace.out","W", stdout); $scanf"%d%d", n,m); s=0; t=2*n+1; the f

http://acm.hdu.edu.cn/showproblem.php?pid=1083Two-figure matching is used a lot.This problem requires only a simplified binary match#include #include#include#defineMAXM 410using namespacestd;intP,n;intMASTER[MAXM];intLINKING[MAXM][MAXM];intHAS[MAXM];intSolveintX//x Lesson{ inti; for(i=1; i) { if(linking[x][i]!Has[i]) {Has[i]=1; if(!master[i]| |solve (Master[i])) {Master[i]=x; return 1; } } } return 0;}intMain () {intm,sum=0; scanf ("%d",m); while(m--) {scanf ("%d%d",p

for(inti =0; I ) About if(Visx[i]) theLx[i]-=D; the for(inti =0; i ) the { + if(Visy[i]) ly[i] + =D; - ElseSlack[i]-=D; the }Bayi } the } the intres =0; - for(inti =0; i ) - if(Linker[i]! =-1) theRes + =g[Linker[i]][i]; the returnRes; the } the - voidINI () the { thescanf"%d",n); the inti,j;94 for(i =0; I ){ the for(j =0; J ) thescanf"%d",g[i][j])

Refer to this blog:http://blog.csdn.net/ascii991/article/details/7466278#include #includestring.h>#include#include#include#includeusing namespaceStd;typedefLong LongLL;Const intN = 1e2+5;intHead[n],tot,p,h[n],n, out[N],inch[N];structedge{intU,v,next;} Edge[n*n],e[n*N];voidAddintUintv) {edge[tot].u=u; EDGE[TOT].V=v; Edge[tot].next=Head[u]; Head[u]=tot++;}voidAddedge (intUintv) {E[P].V=v; E[p].next=H[u]; H[u]=p++;}intClk,dfn[n],low[n],cnt,bel[n];BOOLInstack[n];stackint>s;voidTargin (intu) {Dfn[u]=

The main topic: the direction of the map to cut pointsTopic Ideas:A point u is a cut point when and only if one of the two two conditions is met:1. The point is the root node and has at least two child nodes2.u is not a root, and satisfies the presence (U,V) as a branch edge (or parent-child edge, that is, U is the father of V in the search tree), making DFN (U) Then pay attention to the read in, Easy re#include #includestring.h>#include#include#include#include#include#defineMAXSIZE 1005#defineL