"Codevs1227" Check number 2 (fee flow)

Test instructions: give a matrix of n*n, each with a nonnegative integer Aij, (Aij <=)

Now starting from (a), you can go to the right or down, and finally arrive (N,n), each reached a lattice, the number of the lattice is taken out, the number of the lattice becomes 0,

This takes a total of k times, now requires K times to reach the number of squares and the largest.

n<=50,k<=10

Idea: Cost flow

Splitting each point into an out point and an in point (i,j,1..2), this idea is similar to the maximum flow

(i,j,1) (i,j,2) with two edges:

Capacity is 1, cost a[i,j]

Capacity is K, cost is 0

(i,j,2)-(i+1,j,1) with a capacity of k, the cost of 0 side (i,j+1) Similar

(n,n,2)->t with a flow of k, with a cost of 0 side limit traffic

Run the maximum cost of the maximum flow on the line, in fact, is SPFA triangle inequality change direction

1 varFanArray[1..200000] ofLongint;2Q:Array[0..20000] ofLongint;3Head,vet,next,len1,len2:Array[1..20000] ofLongint;4PreArray[1..20000,1..2] ofLongint;5InqArray[1..20000] ofBoolean;6Dis:Array[1..20000] ofLongint;7A:Array[1.. -,1.. -] ofLongint;8Num:Array[1.. -,1.. -,1..2] ofLongint;9 N,m,k1,i,j,k,tot,s,source,src:longint;Ten Ans:int64; One  A functionmin (x,y:longint): Longint; - begin -  ifX<y Thenexit (x); the exit (y); - End; -  - procedureAdd (a,b,c,d:longint); + begin - Inc (TOT); +next[tot]:=Head[a]; Avet[tot]:=b; atlen1[tot]:=C; -len2[tot]:=D; -head[a]:=tot; -  - Inc (TOT); -next[tot]:=Head[b]; invet[tot]:=A; -len1[tot]:=0; tolen2[tot]:=-D; +head[b]:=tot; - End; the  * functionSpfa:boolean; $ varU,t,w,i,e,v:longint;Panax Notoginseng begin -   fori:=1  toS Do the  begin +dis[i]:=-Maxlongint; Ainq[i]:=false; the  End; +t:=0; w:=1; q[1]:=source; Inq[source]:=true; dis[source]:=0; -   whileT<w Do $  begin $Inc (T); U:=q[tMoD(s+5)]; -inq[u]:=false; -e:=Head[u]; the    whileE<>0  Do -   beginWuyiv:=Vet[e]; the    if(len1[e]>0) and(Dis[u]+len2[e]>dis[v]) Then -    begin WuPre[v,1]:=u; Pre[v,2]:=e; -dis[v]:=dis[u]+Len2[e]; About     if  notINQ[V] Then $     begin -Inc (W); Q[wMoD(s+5)]:=v; -inq[v]:=true; -     End; A    End; +e:=Next[e]; the   End; -  End; $  ifDis[src]=-maxlongint Thenexit (False) the   Elseexit (true); the End; the  the procedureMCF; - varK,e:longint; in T:int64; the begin theK:=SRC; t:=Maxlongint; About   whileK<>source Do the  begin theE:=pre[k,2]; K:=pre[k,1]; thet:=min (t,len1[e]); +  End; -k:=src; the   whileK<>source DoBayi  begin theE:=pre[k,2]; thelen1[e]:=len1[e]-T; -len1[fan[e]]:=len1[fan[e]]+T; -ans:=ans+t*Len2[e]; theK:=pre[k,1]; the  End; the End; the  - begin theAssign (input,'codevs1227.in'); Reset (input); theAssign (output,'Codevs1227.out'); Rewrite (output); the readln (N,K1);94   fori:=1  to 200000  Do the   ifIMoD 2=1  Thenfan[i]:=i+1 the    Elsefan[i]:=i-1; the   fori:=1  toN Do98    forj:=1  toN Doread (a[i,j]); About   fori:=1  toN Do -    forj:=1  toN Do101     fork:=1  to 2  Do102    begin103Inc (s); num[i,j,k]:=s;104    End; the   fori:=1  toN Do106    forj:=1  toN Do107   begin108Add (Num[i,j,1],num[i,j,2],1, A[i,j]);109Add (Num[i,j,1],num[i,j,2],K1,0); the    ifj+1<=n ThenAdd (Num[i,j,2],num[i,j+1,1],K1,0);111    ifi+1<=n ThenAdd (Num[i,j,2],num[i+1J1],K1,0); the   End;113source:=1; src:=s+1; Inc (s); theAdd (Num[n,n,2],SRC,K1,0); the   whileSpfa DoMCF; the writeln (ans);117 close (input);118 close (output);119 End.

"Codevs1227" Check number 2 (fee flow)