poj2135 Farm Tour（费用流）

Description

When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of which contains his house and the Nth of which contains the big barn. A total M (1 <= M <= 10000) paths that connect the fields in various ways. Each path connects two different fields and has a nonzero length smaller than 35,000. 

To show off his farm in the best way, he walks a tour that starts at his house, potentially travels through some fields, and ends at the barn. Later, he returns (potentially through some fields) back to his house again. 

He wants his tour to be as short as possible, however he doesn't want to walk on any given path more than once. Calculate the shortest tour possible. FJ is sure that some tour exists for any given farm.

Input

* Line 1: Two space-separated integers: N and M. 

* Lines 2..M+1: Three space-separated integers that define a path: The starting field, the end field, and the path's length. 

Output

A single line containing the length of the shortest tour. 

Sample Input

4 5
1 2 1
2 3 1
3 4 1
1 3 2
2 4 2

Sample Output

6

Source

USACO 2003 February Gree

 

 

费用流，建图还是比较好想的。据说题中数据范围不对要当成2000个点20000条边才能过。

题目大意及题解：

program rrr(input,output);
const
  inf=987654321;
type
  etype=record
     t,c,w,next,rev:longint;
  end;
var
  e:array[0..40040]of etype;
  a,q,dis,fre,frv:array[0..2020]of longint;
  inq:array[0..2020]of boolean;
  n,m,i,x,y,w,cnt,h,t,ans:longint;
procedure ins(x,y,c,w:longint);
begin
   inc(cnt);e[cnt].t:=y;e[cnt].c:=c;e[cnt].w:=w;e[cnt].next:=a[x];a[x]:=cnt;
end;
procedure add(x,y,w:longint);
begin
   ins(x,y,1,w);e[cnt].rev:=cnt+1;
   ins(y,x,0,-w);e[cnt].rev:=cnt-1;
end;
procedure spfa;
begin
   for i:=1 to n do dis[i]:=inf;
   fillchar(inq,sizeof(inq),false);
   h:=0;t:=1;dis[1]:=0;q[1]:=1;inq[1]:=true;
   while h<>t do
      begin
         inc(h);if h>1000 then h:=1;
         i:=a[q[h]];
         while i<>0 do
            begin
               if (e[i].c>0) and (dis[q[h]]+e[i].w<dis[e[i].t]) then
                  begin
                     dis[e[i].t]:=dis[q[h]]+e[i].w;
                     fre[e[i].t]:=i;frv[e[i].t]:=q[h];
                     if not inq[e[i].t] then
                        begin
                           inc(t);if t>1000 then t:=1;
                           q[t]:=e[i].t;inq[e[i].t]:=true;
                        end;
                  end;
              i:=e[i].next;
            end;
         inq[q[h]]:=false;
      end;
end;
begin
   assign(input,'r.in');assign(output,'r.out');reset(input);rewrite(output);
   readln(n,m);
   fillchar(a,sizeof(a),0);cnt:=0;
   for i:=1 to m do begin readln(x,y,w);add(x,y,w);add(y,x,w); end;
   ans:=0;
   spfa;
   i:=n;while i<>1 do begin e[fre[i]].c:=0;e[e[fre[i]].rev].c:=1;ans:=ans+e[fre[i]].w;i:=frv[i]; end;
   spfa;
   i:=n;while i<>1 do begin ans:=ans+e[fre[i]].w;i:=frv[i]; end;
   write(ans);
   close(input);close(output);
end.