UVa1486 Transportation (minimum cost maximum flow)

The topic probably says that there are N cities, and M has an edge attached to them, each side has two properties one is the AI one is the price that Ci,ai represents the cost of ai*x*x for transporting x units of goods, CI indicates the maximum number of goods that can be transported in the side (<=5). Ask for the minimum cost of transporting K unit cargo from City 1 to City N.

This unit cost is not fixed cost flow is a classic way to build a picture is to split the edge, the problem is split into CI, each unit costs ai*1, ai*3, Ai*5, ai*7. So if only one unit of traffic through this side then the cost is AI, if two units of traffic so the cost is ai*1+ai*3=ai*2*2 ...

Then the problem is solved.

1#include <cstdio>2#include <cstring>3#include <queue>4#include <algorithm>5 using namespacestd;6 #defineMAXN 1117 #defineMAXM 555558 #defineINF (1&LT;&LT;30)9 Ten structedge{ One     intV,cap,cost,next; A }EDGE[MAXM]; - intVS,VT,NV,NE,HEAD[MAXN]; - voidAddedge (intUintVintCapintCost ) { theEdge[ne].v=v; Edge[ne].cap=cap; edge[ne].cost=Cost ; -Edge[ne].next=head[u]; head[u]=ne++; -Edge[ne].v=u; edge[ne].cap=0; edge[ne].cost=-Cost ; -EDGE[NE].NEXT=HEAD[V]; head[v]=ne++; + } -  + intD[MAXN],PRE[MAXN]; A BOOLINQUE[MAXN]; at BOOLSPFA () { -      for(intI=0; i<nv; ++i) { -D[i]=inf; inque[i]=0; -     } -d[vs]=0; inque[vs]=1; -queue<int>que; in Que.push (VS); -      while(!Que.empty ()) { to         intu=Que.front (); Que.pop (); +          for(intI=head[u]; i!=-1; I=Edge[i].next) { -             intv=edge[i].v; the             if(Edge[i].cap && d[v]>d[u]+edge[i].cost) { *d[v]=d[u]+Edge[i].cost; $pre[v]=i;Panax Notoginseng                 if(!Inque[v]) { -inque[v]=1; the Que.push (v); +                 } A             } the         } +inque[u]=0; -     } $     returnd[vt]!=INF; $ } - intMxflow; - intMCMF () { themxflow=0; -     intres=0;Wuyi      while(SPFA ()) { the         intflow=inf,cost=0; -          for(intU=VT; U!=vs; u=edge[pre[u]^1].v) { Wuflow=min (flow,edge[pre[u]].cap); -         } Aboutmxflow+=flow; $          for(intU=VT; U!=vs; u=edge[pre[u]^1].v) { -edge[pre[u]].cap-=flow; -edge[pre[u]^1].cap+=flow; -cost+=Edge[pre[u]].cost; A         } +res+=cost*flow; the     } -     returnRes; $ } the  the intMain () { the     intn,m,k; the      while(~SCANF ("%d%d%d",&n,&m,&k)) { -vs=0; Vt=n; nv=vt+1; Ne=0; inmemset (head,-1,sizeof(head)); theAddedge (VS,1K0); the  About         intu,v,a,c; the          while(m--){ thescanf"%d%d%d%d",&u,&v,&a,&c); the              for(intI=1; i<=c; ++i) { +Addedge (U,v,1, A * (2*i-1)); -             } the         }Bayi         intres=MCMF (); the         if(Mxflow!=k) puts ("-1"); the         Elseprintf"%d\n", res); -     } -     return 0; the}

UVa1486 Transportation (minimum cost maximum flow)