network graph visualization

].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

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

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

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

Title Address: POJ 1144Cut points. There are two inferences to infer whether a point is a cut point:Suppose U is a cut point, and only if the following 1 bars are met1, assuming U is the root, then you must have more than 1 subtrees tree2, suppose you are not a root. Then (U,V) is the branch edge. When Low[v]>=dfn[u].Then according to these two sentences to find a cut point can be.The code is as follows:#include POJ 1144 Network (undirected

If Networkx is not good, pip uninstall then install1) Look at each node's social situationImport Networkx as Nxin [2]: nx.read such as Nx.read_adjlist Nx.read_dot Nx.read_ Edgelist this time should be in a series of edges and Dot folder fbdata in [3]: g = nx.read_edgelist ('0.edges') At this point G inherits the various method of read, viewing the case of G in [5]: Len (G.nodes ()), Len (G.edges ()) out[5]: (333, 2519)The 0 ego user then appears to has 333 friends (nodes), and there are 2519 con

, then you and V belong to a set } Else if(Pre[v] FA) Lowu=min (Lowu, pre[v]); } returnLowu;}intLcaintUintv) { intR =find (U); intL =Find (v); if(r = =l)returnret; if(Pre[u] >Pre[v]) Swap (U, v); while(Pre[u] Pre[v]) { if(Union (pa[v], v)) RET--; V=Pa[v]; } while(U! = V)//v after the previous while either U or the nearest public ancestor of U and v { if(Union (U, Pa[u]) ret--; U=Pa[u]; } returnret;}voidinit () {mem (Pre,0); Mem (PA,0); for(intI=0; ii

]]; $ if(!a[e.to] e.cap >e.flow) { -P[e.to] =G[x][i]; -A[e.to] = min (a[x], E.cap-e.flow); the Q.push (e.to); - }Wuyi } the if(A[t]) Break; - } Wu if(!a[t]) Break; - for(intu = t; U! = S; U = edges[p[u]]. from) { AboutEdges[p[u]].flow + =A[t]; $Edges[p[u] ^1].flow-=A[t]; - } -Flow + =A[t]; - } A returnflow; + } the }; - $ Edmondskarp G; theVectorBak; the

assignment of cows to barns such this no barn ' s capacity is exceeded and the size of the range (i.e ., one more than the positive difference between the The Highest-ranked barn chosen and that lowest-ranked barn chosen) of Barn rankings The cows give their assigned barns is as small as possible.InputLine 1:two space-separated integers, N and BLines 2..n+1:each line contains B space-separated integers which is exactly 1..B sorted into some order. The first integer on line i+1 is the number of