cloudformation template cost

Compared to the maximum flow template: Maximum flow template (Dinic)Paste the minimum fee flow template:const int OO=1E9;CONST int Mm=11111111;const int mn=888888;int node,src,dest,edge;int ver[mm],flow[mm],cost[mm],nex[ Mm];int head[mn],dis[mn],p[mn],q[mn],vis[mn];/** These variables are basically the same as the maximum flow, which increases the

Label: style blog HTTP color Io OS AR for SP Hmm ~ The first write cost flow question... This is the template Question of the billing flow. No more bare questions can be found. Graph creation: Each m (man) serves as the source point, each H (house) serves as the sink point, and each source point and sink point connects one edge separately, the traffic on this edge is 1 (because each source point can only ta

mentioned above, only two-dimensional arrays can be used: when each item can only be retrieved once, the V and U variables adopt a backward loop, when an item is like a full backpack, use a sequential loop. Split an item when there are multiple backpack problems. You may have some ideas here, so you should be able to understand the two-dimensional cost backpack with a look at the code. First, let's look at the complete backpack

1 BOOLSPFA ()2 {3Fill (vs,vs+n+2,false);4Fill (d,d+n+2, INF);5Fill (father,father+n+2,-1);6queueint>Q;7d[0]=0;8Q.push (0);9 while(!q.empty ())Ten { One intu=Q.front (); A Q.pop (); -vs[u]=false; - for(inti=head[u];i!=-1; i=eage[i].next) the { - intv=eage[i].v; - if(eage[i].capd[v]>d[u]+eage[i].cost) - { +d[v]=d[u]+Eage[i].cost; -f

(Heu step 6.1.1) Constructing Roads (Minimum Spanning Tree template question: Minimum Cost for connecting n points), heuconstructing Question: Constructing Roads Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission (s): 207 Accepted Submission (s): 135 Problem DescriptionThere are N versions, which are numbered from

Sample Output179 Sourcekicc Recommendeddy Topic Analysis:Kruscal minimum spanning tree, simple problem. You need to be aware of the following situations:1) Some roads have been repaired by the way: the weight of the edge of the bar is set to 0.MAP[A][B] = Map[b][a] = 0;//for an already existing edge, we set his weight to 0.2) The connection information is converted from a matrix into an edge form.int cnt = 1;for (i = 1; I for (j

(e[i].c>0) and(dis[q[h]]+e[i].w then * begin $dis[e[i].t]:=dis[q[h]]+e[i].w;Panax Notoginsengfrv[e[i].t]:=q[h];fre[e[i].t]:=i; - if notinq[e[i].t] then the begin +Inc (t);ifT> the thent:=1; aq[t]:=e[i].t;inq[e[i].t]:=true; the End; + End; -i:=e[i].next; $ End; $inq[q[h]]:=false; - End; - End; the begin -Assign (input,'r.in'); Assign (output,'R.out'); reset (input); rewrite