tableau network graph

the absence of a setExample: bzoj1497,bzoj2127,bzoj19342. Two points belong to the same set pay Val priceAt this time, the nature of the graph is a binary graph. (Face-beatingThen we can flip the nature of the two-part point, that is, the left point is selected, and the right point even T is selected.So after the transformation the problem became the first one. And just (I,j,val) (J,i,val)Example: bzoj1976

Topic Links:POJ1637Test instructionsA picture that gives a forward and a no-edge, asking if there is a Euler circuit that passes through all sides only onceProblem Solving Ideas:The solution of Euler circuit of mixed graph requires the use of network flow, and the concrete modeling method is as follows:1, first to all the non-directional direction, and then the statistics of all points in the degree and the

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

Transferred from http://blog.csdn.net/regina8023/article/details/45815023 It 's written in front . Network flow with upper and lower bounds is limited by the flow of the edge, which must be within the range of [Down,up]. In fact, the ordinary network flow is a special network with the upper and lower bounds of the flow, but each side of the flow limit to [0,ca

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