Linear Planning and network flow 10 napkin Planning

Algorithm Implementation Question 8-10 napkin plan question (Exercise 8-21)
? Problem description:
In the successive n days of a restaurant, the number of napkins required each day varies. Assume that RI napkin (I = 1,
2 ,..., N ). You can purchase a new napkin at a restaurant. The cost of each napkin is P. Or, you can send the old napkin to the wash area,
It takes M days to wash a piece, and the cost is f points; or it is sent to the slow wash part, it takes n days to wash a piece (n> m), the cost is S <F points.
At the end of each day, the restaurant must decide how many dirty napkins will be delivered to the quick wash department, how many napkins will be delivered to the slow wash department, and how many
Save a few blocks to extend the traffic. However, the sum of the number of napkins washed each day and the number of new napkins purchased must meet the needs of the day.
Design an algorithm to reasonably arrange the napkin use plan for the restaurant in N days to minimize the total cost.
? Programming task:
Program to find an optimal napkin use plan.
? Data input:
Input data is provided by the file input.txt. There are 6 positive integers N, P, M, F, N, S in the row 1st of the file. N is to arrange the napkin
Use the planned number of days; P indicates the cost of each new napkin; M indicates the number of days required for the quick wash part; F indicates the number of days required for the quick wash part.
The cost of a napkin; n is the number of days required to wash a napkin in the slow WASH; s is the cost required to wash a napkin in the slow wash.
The next n rows are the number of napkins required by the restaurant every day in N successive days.
? Result output:
At the end of the program running, output the minimum total cost of using the napkin in the restaurant for N consecutive days to the file output.txt
.
Input File example
Output file example
Input.txt
Output.txt
3 10 2 3 2
5
6
7
145

Question:

In fact, it is not difficult to figure out the diagram. The key lies in the processing of details-edge capacity settings.

In addition, pay attention to reading questions. The question says that the washing can be postponed. Therefore, I should be connected to an edge with a capacity of inf from I to I + 1, because it may have been accumulated for several days.

After this point is noted, similarly, the I + T2 + n connection edge capacity also needs to be INF, and the I + T2 + N capacity is also inf

In fact, this question is a bit like saveprint, and the diagram is clever.

Code:

  1 uses math;  2 const inf=maxlongint;  3 type node=record  4      from,go,next,v,c:longint;  5      end;  6 var e:array[0..2000000] of node;  7     pre,head,q,d:array[0..1000000] of longint;  8     v:array[0..1000000] of boolean;  9     i,j,n,s,t,l,r,mincost,tot,p,t1,t2,p1,p2,x:longint; 10 procedure ins(x,y,z,w:longint); 11  begin 12  inc(tot); 13  with e[tot] do 14   begin 15   from:=x;go:=y;v:=z;c:=w;next:=head[x];head[x]:=tot; 16   end; 17  end; 18 procedure insert(x,y,z,w:longint); 19  begin 20  ins(x,y,z,w);ins(y,x,0,-w); 21  end; 22 function spfa:boolean; 23  var i,x,y:longint; 24  begin 25  fillchar(v,sizeof(v),false); 26  for i:=s to t do d[i]:=inf; 27  l:=0;r:=1;q[1]:=s;d[s]:=0;v[s]:=true; 28  while l<r do 29   begin 30   inc(l); 31   x:=q[l];v[x]:=false; 32   i:=head[x]; 33   while i<>0 do 34    begin 35    y:=e[i].go; 36    if (e[i].v<>0) and (d[x]+e[i].c<d[y]) then 37     begin 38     d[y]:=d[x]+e[i].c; 39     pre[y]:=i; 40     if not(v[y]) then 41      begin 42      v[y]:=true; 43      inc(r); 44      q[r]:=y; 45      end; 46     end; 47    i:=e[i].next; 48   end; 49  end; 50  exit(d[t]<>inf); 51  end; 52 procedure mcf; 53  var i,tmp:longint; 54  begin 55  mincost:=0; 56  while spfa do 57   begin 58   tmp:=inf; 59   i:=pre[t]; 60   while i<>0 do 61    begin 62    tmp:=min(tmp,e[i].v); 63    i:=pre[e[i].from]; 64    end; 65   inc(mincost,tmp*d[t]); 66   i:=pre[t]; 67   while i<>0 do 68    begin 69    dec(e[i].v,tmp); 70    inc(e[i xor 1].v,tmp); 71    i:=pre[e[i].from]; 72    end; 73   end; 74  end; 75 procedure init; 76  begin 77  tot:=1; 78  readln(n,p,t1,p1,t2,p2); 79  s:=0;t:=2*n+1; 80  for i:=1 to n do 81   begin 82   readln(x); 83   insert(s,i,x,0); 84   insert(s,i+n,x,p); 85   insert(i+n,t,x,0); 86   if i+1<=n then insert(i,i+1,inf,0); 87   if i+t1<=n then insert(i,i+t1+n,inf,p1); 88   if i+t2<=n then insert(i,i+t2+n,inf,p2); 89   end; 90  end; 91 procedure main; 92  begin 93  mincost:=0; 94  mcf; 95  writeln(mincost); 96  end; 97 begin 98  assign(input,‘input.txt‘);assign(output,‘output.txt‘); 99  reset(input);rewrite(output);100  init;101  main;102  close(input);close(output);103 end.   

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