upper and lower bounds calculator

otherwise.Any pair of nodes can be connected only by one pipe. if there is a pipe from node A to Node B, then there is no pipe from B to. not a single node is connected with itself.There is no pipe which connects nodes number 1 and N. substance can flow only from the beginning of a pipe to its end.OutputWrite one integer number in the first line of the output-it ought to be the minimal speed of the producing substance in the first node.Write M integers in the second line-I-th number ought to be

Topic Links:zoj1994Title Description:A budget is now set for a multi-zone contest, which is a matrix that represents the different kinds of expenses and represents the expenses of different divisions. The organizing committee has held a meeting to discuss the sum of all kinds of expenses, as well as the total cost of each division. In addition, the organizing Committee discussed a number of special constraints: for example, it was suggested that the computer centre needed at least 1000K (Iranian

"Author: Mengxiang Month Blog: Http://blog.csdn.net/mengxiangyue" In writing a program, we want a function to receive parameters, there is a range of limitations, this time we can consider the use of generic upper and lower bounds to solve. Let's look at a piece of code first and then explain it in detail. Class pointIn this program we define a function show, wh

,pre[u]); Pre[u]=tot++; E[tot]=edge (V,u,0, Pre[v]); PRE[V]=tot++;}voidinit () {memset (inch,0,sizeof(inch)); memset ( out,0,sizeof( out)); memset (PRE,-1,sizeof(pre)); Tot=0;}/*******************dinic************************/intBFs () {intHead=0, tail=1; memset (H,-1,sizeof(h)); q[0]=vs;h[vs]=0; while(head!=tail) { intu=q[head++]; for(intI=pre[u];~i;i=E[i].next) { intv=e[i].v,w=E[I].W; if(wh[v]==-1) {H[v]=h[u]+1; Q[tail++]=v; } } } returnh[vt]!=-1;}intDfsintUint

POJ 2396 determining feasible solutions of the upper and lower bounds of the active sink Question: there is a series of n rows and m columns. The values and values of each element and value in each column are given to you. The following are restrictions on element values, for example, 0 0> 1 indicates that all elements in the series are greater than 1 and 0 indi

the Company A wants to transport as many goods as possible from the city S to City T. So they ask company B to do the transportation. There are n cities the problem and There are directed from roads I to City J with cities capacity. After a long negotiation, a and B reach a consensus that a A and a-decide to transport the goods (of course under the CO Ndition that most goods are transported and then the quantity transported in each road cannot exceed the capacity) and the N B can assign non-neg

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

Network stream with the upper and lower bounds of ZOJ 2314 Given m and n nodes, the minimum traffic and the maximum traffic on each side are equal to the inbound and outbound traffic on each node. Can this problem be solved, you can output the traffic of each edge. Idea: a network stream with the upper and

"Topic link"http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=11025ModelThere is a minimum flow of Yuanhui points in the upper and lower bounds. The minimum flow that satisfies both the upper and lower bounds and the flo

(); mark[now]=0; for(intI=head[now]; I I=edge[i].next)if(edge[i^1].cap dis[edge[i].to]>dis[now]+edge[i^1].cost) {dis[edge[i].to]=dis[now]+edge[i^1].cost; if(!mark[edge[i].to]) Q.push (edge[i].to), mark[edge[i].to]=1; } } returndis[s]!=inf;}intDfsintLocintLow ) {Mark[loc]=1; if(loc==t)returnLow ; intW,used=0; for(intI=head[loc]; I I=edge[i].next)if(Edge[i].cap !mark[edge[i].to] dis[edge[i].to]==dis[loc]-edge[i].cost) {W=dfs (Edge[i].to,min (low-used,edge[i].cap)); Edge[i].cap-=w; e

); J:=Next[j]; End Exit (true); end;function Solve:boolean;vari,sum:longint;begin sum:=0; whileSpfa DoInc (Sum,dfs (s),1000000007)); if(not Jud) then exit (false)ElseExittrue); End;begin readln (n); ans:=0; fori:=1to n Dobegin forj:=1to n Dobegin Read (A[i,j]); Inc (Ans,trunc (A[I,J)); end; READLN; End Lx:=0; rx:=10000000007; sum:=-1; whileLx Dobegin Mid:= (LX+RX) >>1; Fillchar (Next,sizeof(next),0); Fillchar (Link,sizeof(link),0); Fillchar (INP,sizeof(INP),0); E:=0; s:=0; t:=2*n+1;

minimum residual amount of all arcs so far{ if(x = = T | | a = =0)returnA; int i = cur[x];//Dfs to the same point multiple times when backtracking if(i = =0) i =Head[x]; intFlow =0, F; for(; i;i = E[i]. Next) {//starting with the last thought of the arc intv =e[i].to; if(D[v] = = d[x]+1 (f = DFS (V,min (A,e[i].cap-e[i].flow))) >0) {E[i].flow+=F; E[i^1].flow-=F; Flow+=F; A-= f;//Residual quantity-flow if(A = =0) Break; } } returnflow;}intDinic () {intFlow =0; whil

Click to open linkTest instructions: given M-bar and N-nodes, each edge of the least traffic and the most traffic, to ensure that each node's access to traffic and equality, ask can be formed, you can output each side of the trafficIdea: A network flow with the upper and lower bounds, because there is the nether, we must run each side of the traffic greater than

Network stream with Upper and Lower Bounds In the past few days, I have read Zhou Yuan's "a simple method to solve Network Flow Problems with the upper and lower limits of traffic" and completed sgu 194 zoj 2314 Reactor Cooling, sgu 176 flow construction and hoj

[Cpp]*Hdu 3157Poj 3801A circuit board with N wiring bars (Number 1 ~ N) There are two power supply terminals +-Then the terminal and minimum current of M parts are given.Find an impossible output if the total current that can make all parts work normally is notIt is actually a problem of having a source sink with the upper and lower bounds of the smallest stream.

"Topic link"http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=20757Test instructionsThe feasible flow of passive and non-sinks with the upper and lower bounds of capacity is obtained.IdeasThe passive non-sink feasible flow requires that all vertices meet the flow balance.The basic idea is to convert to the maximum flow to do.For edges (u,v,b,c), The Edg

sgu194:No Yuanhui upper and lower bounds network flow, to find the maximum flow./*Each side add (U,v,up[i]-dn[i])Each point out[i]=∑dn[i,v];in[i]=∑dn[v,i];Each point in[i]-out[i]>0 Add (S,i, in[i]-out[i]), otherwise add (I,t,out[i]-in[i]);Determine if the side connected to St is full stream.*/#include 　　Upper and

Test instructions Give a matrix of each row and each column and, in the constraints given to some of the rows or points, seek a satisfying matrix. Analysis: to the upper and lower bounds of the network flow, note is equal to an upper and lower

(V, u, G[v],0); inG[V] =Edtot; - } to + intLEVEL[MAXN], S, t; - BOOLBFs () { the Static intQUE[MAXN];intQh0), QT (0); * CLR (level); $LEVEL[QUE[++QT] = s] =1;Panax Notoginseng while(QH! =qt) { - intx = Que[++qh];if(QH = = MAXN-1) QH =0; the for(inti = g[x]; I i = e[i].nx)if(E[I].C !)LEVEL[E[I].V]) { +LEVEL[QUE[++QT] = e[i].v] = level[x] +1; A if(qt = = MAXN-1) qt =0; the } + } - return!!Level[t]; $ } $ intDfsintUintRM) { - if(U = = t)re

Zoj 3229 shoot the bullet Link: http://acm.zju.edu.cn/onlinejudge/showProblem.do? Problemcode = 3229. Question:A diaosi takes a photo of the goddess M. It is planned to take a photo for N days. Each day, the diaosi can take a photo of the goddess C at most, and the number of pictures cannot exceed D every day, in addition, the number of photos taken for each goddess is limited to [Li, Ri]. The total number of photos taken by each goddess for N days cannot exceed GI, and how many photos should b