NOI2006最大获利

终于搞明白了……

x到y有边意味着选x必须选y，所以才会有闭合子图中一个点的后继一定也在这个闭合子图中。

接下来按照s连正权，负权连t就ok了

type node=record
     go,next,c:longint;
     end;
var i,n,m,s,t,max,tmp,ans,tot,x,y,z:longint;
    h,first,q,cur:array[0..100000] of longint;
    e:array[0..500000] of node;
function min(x,y:longint):longint;
 begin
   if x<y then exit(x) else exit(y);
 end;
procedure insert(x,y,z:longint);
 begin
   inc(tot);
   e[tot].go:=y;
   e[tot].c:=z;
   e[tot].next:=first[x];
   first[x]:=tot;
   inc(tot);
   e[tot].go:=x;
   e[tot].c:=0;
   e[tot].next:=first[y];
   first[y]:=tot;
 end;
procedure init;
 begin
   readln(n,m);
   s:=0;t:=m+n+1;tot:=1;
   for i:=1 to n do
    begin
      read(x);
      insert(s,i,x);
    end;
   max:=0;
   for i:=1 to m do
    begin
      readln(x,y,z);inc(max,z);
      insert(n+i,t,z);
      insert(x,n+i,maxlongint>>2);
      insert(y,n+i,maxlongint>>2);
    end;
 end;
function bfs:boolean;
 var i,x,y,head,tail:longint;
 begin
   fillchar(h,sizeof(h),0);
   head:=0;tail:=1;q[1]:=s;h[s]:=1;
   while head<tail do
    begin
      inc(head);
      x:=q[head];
      i:=first[x];
      while i<>0 do
        begin
          y:=e[i].go;
          if (h[y]=0) and (e[i].c<>0) then
           begin
             h[y]:=h[x]+1;
             inc(tail);
             q[tail]:=y;
           end;
          i:=e[i].next;
        end;
    end;
   exit(h[t]<>0);
 end;
function dfs(x,f:longint):longint;
 var i,tmp,used:longint;
   begin
     if (x=t) or (f=0) then exit(f);
     tmp:=0;used:=0;
     i:=cur[x];
     while i<>0 do
       begin
         y:=e[i].go;
         if (e[i].c<>0) and (h[y]=h[x]+1) then
           begin
             tmp:=dfs(y,min(e[i].c,f-used));
             dec(e[i].c,tmp);
             inc(e[i xor 1].c,tmp);
             if e[i].c<>0 then cur[x]:=i;
             inc(used,tmp);
             if used=f then exit(f);
           end;
         i:=e[i].next;
       end;
     if used=0 then h[x]:=-1;
     exit(used);
   end;
procedure dinic;
 begin
   ans:=0;
   while bfs do
     begin
       for i:=0 to t do cur[i]:=first[i];
       inc(ans,dfs(0,maxlongint>>2));
     end;
 end;
procedure main;
 begin
   dinic;
   writeln(max-ans);
 end;
begin
  init;
 main;
end.