UVa10779 collectors problem (maximum flow)

It is easy to think of the source point to the type has a sticker edge, the capacity of Bob at the beginning of the quantity, and then the sticker to the meeting point edge, the capacity of 1.

The next part is the exchange of the edge. Note that the swap takes place one at a time.

Swap, is for both stickers, only occurs in one sticker a friend has a quantity of 0, while another friend has more than 1.

So the map, for each friend, if the sticker number is 0, then this sticker to this friend with a capacity of 1 side; If the number x is greater than 1, the friend over there has a capacity of x-1 on the side of the sticker.

1#include <cstdio>2#include <cstring>3#include <queue>4#include <algorithm>5 using namespacestd;6 #defineINF (1&LT;&LT;30)7 #defineMAXN 558 #defineMAXM 55*55*29 Ten structedge{ One     intV,cap,flow,next; A }EDGE[MAXM]; - intVs,vt,ne,nv; - intHEAD[MAXN]; the  - voidAddedge (intUintVintcap) { -Edge[ne].v=v; Edge[ne].cap=cap; edge[ne].flow=0; -Edge[ne].next=head[u]; head[u]=ne++; +Edge[ne].v=u; edge[ne].cap=0; edge[ne].flow=0; -EDGE[NE].NEXT=HEAD[V]; head[v]=ne++; + } A  at intLEVEL[MAXN]; - intGAP[MAXN]; - voidBFs () { -memset (level,-1,sizeof(level)); -Memset (Gap,0,sizeof(GAP)); -level[vt]=0; ingap[level[vt]]++; -queue<int>que; to Que.push (VT); +      while(!Que.empty ()) { -         intu=Que.front (); Que.pop (); the          for(intI=head[u]; i!=-1; I=Edge[i].next) { *             intv=edge[i].v; $             if(level[v]!=-1)Continue;Panax Notoginsenglevel[v]=level[u]+1; -gap[level[v]]++; the Que.push (v); +         } A     } the } +  - intPRE[MAXN]; $ intCUR[MAXN]; $ intIsap () { - BFS (); -memset (pre,-1,sizeof(pre)); thememcpy (Cur,head,sizeof(head)); -     intu=pre[vs]=vs,flow=0, aug=INF;Wuyigap[0]=NV; the      while(level[vs]<NV) { -         BOOLflag=false; Wu          for(int&i=cur[u]; i!=-1; I=Edge[i].next) { -             intv=edge[i].v; About             if(Edge[i].cap!=edge[i].flow && level[u]==level[v]+1){ $flag=true; -pre[v]=u; -u=v; -                 //aug= (Aug==-1?edge[i].cap:min (Aug,edge[i].cap)); AAug=min (aug,edge[i].cap-edge[i].flow); +                 if(v==VT) { theflow+=; -                      for(u=pre[v]; V!=vs; v=u,u=Pre[u]) { $edge[cur[u]].flow+=; theedge[cur[u]^1].flow-=; the                     } the                     //Aug=-1; theaug=INF; -                 } in                  Break; the             } the         } About         if(flag)Continue; the         intMinlevel=NV; the          for(intI=head[u]; i!=-1; I=Edge[i].next) { the             intv=edge[i].v; +             if(Edge[i].cap!=edge[i].flow && level[v]<minlevel) { -Minlevel=Level[v]; thecur[u]=i;Bayi             } the         } the         if(--gap[level[u]]==0) Break; -level[u]=minlevel+1; -gap[level[u]]++; theu=Pre[u]; the     } the     returnflow; the } -  the intMain () { the     intt,n,m,a,b; thescanf"%d",&t);94      for(intCse=1; cse<=t;++CSE) { thescanf"%d%d",&n,&m); thememset (head,-1,sizeof(head)); thevs=0; vt=n+m+1; nv=vt+1; Ne=0;98  Aboutscanf"%d",&a); -         intcnt[ -]={0};101          while(a--){102scanf"%d",&b);103++Cnt[b];104         } the          for(intI=1; i<=m;++i) {106             if(cnt[i]==0)Continue;107 Addedge (Vs,i,cnt[i]);108         }109          the          for(intI=1; i<=m;++i) Addedge (I,VT,1);111          the          for(intI=2; i<=n;++i) {113scanf"%d",&a); the             intcnt[ -]={0}; the              while(a--){ thescanf"%d",&b);117++Cnt[b];118             }119              for(intj=1; j<=m;++j) { -                 if(cnt[j]>=2){121Addedge (i+m,j,cnt[j]-1);122}Else if(cnt[j]==0){123Addedge (J,i+m,1);124                 } the             }126         }127          -printf"Case #%d:%d\n", Cse,isap ());129     } the     return 0;131}

UVa10779 collectors problem (maximum flow)