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The least cost feasible flow is obtained by the method of calculating the feasible flow with the nether network flow.Write the cost stream multi-augmented, a little faster than bare ek.In fact, this problem has a very good method of building maps, but it does not.#include 　　bzoj3876: [Ahoi2014] regional plot fee flow

if(!Now ) +d[u[i]]=0; thew[i]-=Now ; -w[i^1]+=Now ; $rest-=Now ; the } the returnF-rest; the } the voidJianDoublemi) - { inCnt=1; theMemset (Head,0,sizeof(int) * (t+1)); the for(intI=1; i) About if(a[i].zmi) the Jia (a[i].x,a[i].y,a[i].z); the Else the Jia (a[i].x,a[i].y,mi); + return; - } the intMain ()Bayi { thescanf"%D%D%LF",n,m,p); the for(intI=1; i) - { -scanf"%D%D%LF",a[i].x,a[i].y,a[i].z); the Jia (a[i].x,a[i].y,a[i].z); the

,j+1), K,0); }}intD[n],q[n],head,tail,inq[n],pre[n],pos[n];inlinevoidLopintAMP;X) {if(x==n) x=1;}BOOLSPFA () {memset (d,127,sizeof(d)); memset (INQ,0,sizeof(INQ)); Head=tail=1; D[s]=0; inq[s]=1; q[tail++]=s; Pre[t]=-1; while(head!=tail) { intu=q[head++];inq[u]=0; Lop (head); for(intI=h[u];i;i=e[i].ne) { intv=e[i].v,w=E[I].W; if(d[v]>d[u]+we[i].c>e[i].f) {D[v]=d[u]+W; PRE[V]=u;pos[v]=i; if(!inq[v]) q[tail++]=v,inq[v]=1, lop (tail); } } } returnpre[t]!=-1;}intMCMF

date. With decades of operating system development, operating system maintenance, operating system education experience, this is the most worthy of participation in RHCE certification learning place.3, to learn. RHCE certification for students to bring the new technology concept, in fact, is also to guide students to learn, to learn the most advanced technology in the IT industry, to continue to explore the future development of IT industry, in the IT field to go further. There is no forever te

Click to open linkTest instructions: There are n individuals and M to the topic, each person to do the probability of a matrix given, ask how to allocate can make the probability of the largest, there is a limit is to do so far every two people do the number of problems of the gap can not exceed 1, that is, the first n road problems, must be a person to do aIdeas: Online are more than a bit of DP, with the network flow can also, but a lot of trouble, but the weak is a little DP will not be a con

The first question of this question is a bare maximum flow, not much to say, the key lies in the second question. First there is a conclusion that Bob must add the cost to a side, so we can two points per side of the flow, to verify whether the maximum flow can beCode#include Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced. "Sdoi 2013" "Bzoj 3130" fee flow

! = S; U = edges[pre[u]].from) {Edges[pre[u]].flow + = a[t]; edges[pre[u]^1].flow-= a[t]; }return 1;}intMCMF (intSintT, ll cost) {ll flow =0; Cost =0; while(BF (S, t, flow, cost));returnFlow;}voidInput () {s =0, t = n + M +3;intA, B; for(inti =1; I scanf("%d", a); for(intj =0; J scanf("%d", b); Wk[i].push_back (b); } } for(inti =1; I scanf("%d", rec[i]); }}voidBuild () { for(inti =1; I 1,0); for(inti =1; I 1,0); for(inti =1; I for(intj =0; J if(Wk[i][j] = = Rec[i]) Addedge (i,

Portal: Click to open linkTest instructions: to n points and M Bar has a forward edge, to find out a number of rings out, each ring point at least 2, all points are to be covered 1 times, and only 1 times. Ask the sum of the lengths of all ringsThis problem can also be done in km, here is the practice of building the cost streamFor this problem, it is very wonderful to build the map.Since each point is in the 1, the degree is 1So I would think of splitting each point into 2 points, denoted by I

; } } } }if(d[t]==-1)return 0; FLOW+=A[T]; COST+=D[T]*A[T]; u=t; while(u!=s) {edges[p[u]].flow+=a[t]; edges[p[u]^1].FLOW-=A[T]; U=edges[p[u]].from; }return 1;}intMincost (intSintT) {intflow=0, cost=0; while(Bellmanford (S,t,flow,cost));returnCost;}intMain () {intI,j; while(~scanf("%d%d", n,k)) { for(i=0; i2*n*+1; i++) g[i].clear (); Edges.clear (); for(i=1; i for(j=1; jscanf("%d", c);intp= (I-1) *n+j;intQ=p+n*n; Addedge (P,q,1, c); Addedge (P,q,k,0);if

=0; while(SPFA (s,t)) {intMin =INF; for(inti = pre[t];i! =-1; i = pre[edge[i^1].to]) {if(Min > Edge[i].cap-Edge[i].flow) Min= Edge[i].cap-Edge[i].flow; } for(inti = pre[t];i! =-1; i = pre[edge[i^1].to]) {Edge[i].flow+=Min; Edge[i^1].flow-=Min; Cost+ = edge[i].cost*Min; } Flow+=Min; } returnflow;}intn,m,k;Charstr[ A][ A];voidsolve () {init (2*n*m +3); intStart =2*n*m; intEnd =2*n*m+2; Addedge (Start,start+1K0); for(inti =0; I ) for(intj =0; J ) {Addedge (Start,2* (I*M+J),1,0

according cost Information Network understand that the Guangdong Development and Reform Commission released yesterday on the construction Project Cost Consulting Service charge of the reply, the contents of the notice copied to the Municipal price Bureau, Shenzhen Development and Reform Commission , Market Supervision Bureau, Foshan Shunde District Development Planning and Statistics Bureau. Notice the specific content as shown in the file:650) this.width=650; "src=" Http://info.zjtcn.com/uploa

units of goods, output-1.Sample Input2 1 21 2 1 22 1 21 2 1 12 2 21 2 1 21 2 2 2Sample Output4-13Test instructionsThere are n cities and m one-way routes. For each line, there are four messages: The starting point A, the end point B, the cost factor C, the capacity D (1Analytical:This question is to use the cost flow to write, but the premise of the cost flow is proportional to the cost and flow, but the cost of the problem and the flow of the square is proportional. So here's the conversion, t

each point with a different dish with a capacity of 1, the cost of t[i][j]xk side (k means the countdown to do, because the last to do only this person need to wait, do this dish time will not be other dishes), but we found that in fact these sides must have a lot of will not go to.So we just have to start with the countdown to the first person, that is t[i][j] even good.Each time the augmentation, in fact, as long as one is to do the rest of the dish is to withdraw vegetables, so we need to fi

[fa[p],p];//build reverse Side 1 A[fa[p],p]:=a[fa[p],p]-min; a[p,fa[p]]:=a[p,fa[p]]+min;//build reverse Side 2 p:=Fa[p]; End; Sum:=sum+d[t];//Add the cost of this augmentation, update the answer exit (min);End;beginread (n,m); fori:=1 toN Do begina[0, i]:=1; c[0, i]:=0;//build a diagram 1End; fori:=1 toM Do beginA[i+n,n+m+1]:=1; c[i+n,n+m+1]:=0;//build a diagram 2End; fori:=1 toN Do forj:=1 toM Do beginread (k); A[i,n+j]:=1; c[i,n+j]:=k;//Build a diagram 3End; N:=n+m+1; sum:=0; k:=1;

Topic: Given a sequence of length n, it is required to select some number so that any one length is the maximum number of k in a m interval, and the maximum andThe charge flow is just running.This sequence is connected to a line with a flow of K cost of 0, and then the point I points to the i+m point with a cost of a[i] the edge of 1 flowMaximum cost of running the maximum flow can beCard Simple Type differential evaluation ....#include Bzoj 1283 sequence fe