3891: [Usaco2014 Dec]Piggy Back

3891: [Usaco2014 Dec]Piggy Back

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 116  Solved: 92
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Description

Bessie and her sister Elsie graze in different fields during the day, and in the evening they both want to walk back to the barn to rest. Being clever bovines, they come up with a plan to minimize the total amount of energy they both spend while walking. Bessie spends B units of energy when walking from a field to an adjacent field, and Elsie spends E units of energy when she walks to an adjacent field. However, if Bessie and Elsie are together in the same field, Bessie can carry Elsie on her shoulders and both can move to an adjacent field while spending only P units of energy (where P might be considerably less than B+E, the amount Bessie and Elsie would have spent individually walking to the adjacent field). If P is very small, the most energy-efficient solution may involve Bessie and Elsie traveling to a common meeting field, then traveling together piggyback for the rest of the journey to the barn. Of course, if P is large, it may still make the most sense for Bessie and Elsie to travel separately. On a side note, Bessie and Elsie are both unhappy with the term "piggyback", as they don't see why the pigs on the farm should deserve all the credit for this remarkable form of transportation. Given B, E, and P, as well as the layout of the farm, please compute the minimum amount of energy required for Bessie and Elsie to reach the barn.

给定一个N个点M条边的无向图，其中Bessie在1号点，Elsie在2号点，它们的目的地为N号点。Bessie每经过一条边需要消耗B点能量，Elsie每经过一条边需要消耗E点能量。当它们相遇时，它们可以一起行走，此时它们每经过一条边需要消耗P点能量。求它们两个到达N号点时最少消耗多少能量？

Input

The first line of input contains the positive integers B, E, P, N, and M. All of these are at most 40,000. B, E, and P are described above. N is the number of fields in the farm (numbered 1..N, where N >= 3), and M is the number of connections between fields. Bessie and Elsie start in fields 1 and 2, respectively. The barn resides in field N. The next M lines in the input each describe a connection between a pair of different fields, specified by the integer indices of the two fields. Connections are bi-directional. It is always possible to travel from field 1 to field N, and field 2 to field N, along a series of such connections.

Output

A single integer specifying the minimum amount of energy Bessie and

Elsie collectively need to spend to reach the barn.  In the example

shown here, Bessie travels from 1 to 4 and Elsie travels from 2 to 3

to 4.  Then, they travel together from 4 to 7 to 8.

 

Sample Input

4 4 5 8 8
1 4
2 3
3 4
4 7
2 5
5 6
6 8
7 8

Sample Output

22

HINT

 

Source

Silver

题解：直接求出1、2、n点到各点的距离（由于是无向图所以方向神马的直接无视之），然后枚举各个汇合点，然后计算各个点的代价，然后输出，然后AC

一开始在怀疑这样子是否一定可行，是否会存在两者会合后再次分开的可能，但实际上，如果两个人一起行动更合适的话，那么相遇就不需要再分离；如果两个人不适合一起行动的话，那么干脆就不需要相遇。所以不要相遇了再分离，综上。（HansBug：嗨。。这话写的。。。为啥感觉越读越戳泪点TT）

（PS：程序里面我很逗比的还弄了个b作为反向map，实际上完全不必，一开始我没发现这个是无向图，所以b用来存储反向图，后来才发现我想多了TT）

 /**************************************************************
     Problem:
     User: HansBug
     Language: Pascal
     Result: Accepted
     Time: ms
     Memory: kb
 ****************************************************************/

 type
     point=^node;
     node=record
                g,w:longint;
                next:point;
     end;
     map=array[..] of point;
     arr=array[..] of longint;
 var
    i,j,k,l,m,n,a1,a2,a3:longint;
    a,b:map;
    c,e,f,g:arr;
    d:array[..] of longint;
 function min(x,y:longint):longint;inline;
          begin
               if x<y then min:=x else min:=y;
          end;
 function max(x,y:longint):longint;inline;
          begin
               if x>y then max:=x else max:=y;
          end;
 procedure add(x,y,z:longint;var a:map);inline;
           var p:point;
           begin
                new(p);p^.g:=y;p^.w:=z;
                p^.next:=a[x];a[x]:=p;
           end;
 procedure spfa(x:longint;a:map;var c:arr);inline;
           var f,r:longint;p:point;
           begin
                fillchar(g,sizeof(g),);
                fillchar(c,sizeof(c),);
                f:=;r:=;d[]:=x;g[x]:=;c[x]:=;
                while f<r do
                      begin
                           p:=a[d[f]];
                           while p<>nil do
                                 begin
                                      if (c[p^.g]=) or((c[p^.g]>) and (c[p^.g]>(c[d[f]]+p^.w))) then
                                         begin
                                              c[p^.g]:=c[d[f]]+p^.w;
                                              if g[p^.g]= then
                                                 begin
                                                      g[p^.g]:=;
                                                      d[r]:=p^.g;
                                                      inc(r);
                                                 end;
                                         end;
                                      p:=p^.next;
                                 end;
                           g[d[f]]:=;
                           inc(f);
                      end;
                for i:= to n do dec(c[i]);
           end;

 begin
      readln(a1,a2,a3,n,m);
      for i:= to n do a[i]:=nil;
      for i:= to n do b[i]:=nil;
      for i:= to m do
          begin
               readln(j,k);
               add(j,k,,a);
               add(k,j,,a);
          end;
      spfa(,a,c);
      spfa(,a,e);
      spfa(n,a,f);
      l:=maxlongint;
      for i:= to n do
          if (c[i]<>-) and (e[i]<>-) and (f[i]<>-) then
             l:=min(l,a1*c[i]+a2*e[i]+a3*f[i]);
      writeln(l);
 end.