"Codevs1034" Homeland (max Flow, split point)

Test instructions: Because of the insane destruction of nature, people realize that after about 2,300 years, the earth can no longer live, and then set up new green space on the moon so as to emigrate when needed. Surprisingly, the 2177 winter due to unknown reasons, the Earth's environment has been a chain collapse, the human must in the shortest possible time to move inland to the moon.
The existing n space stations are between the Earth and the Moon (ref. 1). n), M public transport spacecraft shuttling back and forth, each space station SI can accommodate unlimited people, each spacecraft pi can only accommodate the HPI people. For each spacecraft Pi, a series of space stations will be periodically docked (Si1,si2 ... Sir), such as: (1,3,4) means docking station 134134134.... It takes 1 for any spaceship to sail from any space station to another arbitrary space station. One can only ship or disembark when a spaceship docks (or the Earth or the moon) at a space station. The initial people were all on earth, and the spacecraft was all at the initial station (the spacecraft Pi was in Si1), with the goal of allowing all people to move all the way to the moon as soon as possible.

The first act of the file is three positive integers n (number of space stations), m (number of spacecraft), K (number of people on earth that need to be transported), of which 1<=m<=13, 1<=n<=20, 1<=k<=50.

Idea: Run the maximum flow by the time split point

Num[i,j] Represents the J point after I seconds

(I,J)--(I+1,J) even with unlimited flow

(i,x)--(i+1,y) with a flow of h[k], X, Y is a planet where the K spacecraft stay at the i,i+1 time, respectively.

Enumerate the answers and add edges, dinic

Note that the maximum flow after each run to restore the flow of reconstruction, changed for a long time did not change out

I'm still too simple.

By the way, 1999 years of official data, the second group is wrong, the correct answer is that 5,out is 7.

1 varHead,vet,next,len,dis,gap,fan,h,r,save:Array[0..100000] ofLongint;2A,b,num:Array[0.. the,0.. the] ofLongint;3 N,m,i,j,k,x,ans,tot,source,src,s:longint;4 5 procedureAdd (a,b,c:longint);6 begin7 Inc (TOT);8next[tot]:=Head[a];9vet[tot]:=b;Tenlen[tot]:=C; Onehead[a]:=tot; A End; -  - functionmin (x,y:longint): Longint; the begin -  ifX<y Thenexit (x); - exit (y); - End; +  - functionDFS (u,aug:longint): Longint; + varE,v,val,flow,t:longint; A begin at  ifU=src Thenexit (); -flow:=0; -E:=head[u]; val:=s-1; -   whileE<>0  Do -  begin -v:=Vet[e]; in   ifLen[e]>0  Then -   begin to    ifdis[u]=dis[v]+1  Then +    begin -T:=dfs (V,min (len[e],aug-flow)); thelen[e]:=len[e]-T; *len[fan[e]]:=len[fan[e]]+T; $flow:=flow+T;Panax Notoginseng     ifDis[source]>=s Thenexit (flow); -     ifAug=flow ThenBreak ; the    End; +val:=min (val,dis[v]); A   End; thee:=Next[e]; +  End; -  ifflow=0  Then $  begin $ Dec (Gap[dis[u]]); -   ifgap[dis[u]]=0  Thendis[source]:=s; -dis[u]:=val+1; the Inc (Gap[dis[u]); -  End;Wuyi exit (flow); the End; -  Wu functionMaxflow:longint; - varAns:longint; About begin $Fillchar (dis,sizeof (DIS),0); -Fillchar (Gap,sizeof (GAP),0); -gap[0]:=s; -ans:=0; A   whileDis[source]<s Doans:=ans+DFS (source,maxlongint); + exit (ans); the End; -  $ begin theAssign (input,'codevs1034.in'); Reset (input); theAssign (output,'Codevs1034.out'); Rewrite (output); the readln (n,m,k); the   fori:=0  to 101  Do -    forj:=1  ton+2  Do in   begin theInc (s); num[i,j]:=s; the   End; About   fori:=1  toM Do the  begin the read (h[i]); Read (R[i]); the    forj:=1  toR[i] Do +   begin - read (a[i,j]); the    ifa[i,j]=0  Thena[i,j]:=n+1;Bayi    ifa[i,j]=-1  Thena[i,j]:=n+2; the   End; thex:=1; B[i,0]:=a[i,1]; -    forj:=1  to 101  Do -   begin theInc (X);ifx=r[i]+1  Thenx:=1; theb[i,j]:=A[i,x]; the   End; the  -  End; theans:=-1; thesource:=num[0, n+1]; the 94   fori:=0  to 101  Do the  begin the    forj:=1  toTot DoSAVE[J]:=LEN[J]; //It is extremely important to reconstruct the capacity of the edge thesrc:=num[i,n+2]; s:=src;98   ifMaxflow>=k Then beginAns:=i; BreakEnd; About    forj:=1  toTot DoLEN[J]:=SAVE[J]; //It is extremely important to reconstruct the capacity of the edge -    forj:=1  toM Do101   begin102fan[tot+1]:=tot+2;103fan[tot+2]:=tot+1;104Add (num[i,b[j,i]],num[i+1, b[j,i+1]],h[j]); theWriteln (I,' ', B[j,i],' ', i+1,' ', b[j,i+1],' ', H[j]);106Add (num[i+1, b[j,i+1]],num[i,b[j,i]],0);107Writeln (i+1,' ', b[j,i+1],' 'I' ', B[j,i],' ',0);108   End;109    forj:=1  ton+2  Do the   begin111fan[tot+1]:=tot+2; thefan[tot+2]:=tot+1;113Add (num[i,j],num[i+1, j],maxlongint); theWriteln (I,' 'J' ', i+1,' 'J' ', maxlongint); theAdd (num[i+1, J],num[i,j],0); theWriteln (i+1,' 'J' 'I' 'J' ',0);117   End;118 Writeln;119  End; -  ifAns>0  Thenwriteln (ANS)121   ElseWriteln (0);122//close (input);123//close (output);124 End.

"Codevs1034" Homeland (max Flow, split point)