bzoj1570

购买的机票限制和数据范围很容易想到是网络流
不难想到每个城市按时刻拆点，这也是一个经典模型
由于时间不会太大，我们穷举时间，不断在残留网络上建图，跑最大流
直至总流量为k即可

 const inf=;
 type node=record
        po,next,flow:longint;
      end;

 var e:array[..] of node;
     a,b,c,p,numh,h,d,cur,pre:array[..] of longint;
     s,max,t,ans,i,n,m,k,len:longint;

 function min(a,b:longint):longint;
   begin
     if a>b then exit(b) else exit(a);
   end;

 procedure add(x,y,f:longint);
   begin
     inc(len);
     e[len].po:=y;
     e[len].flow:=f;
     e[len].next:=p[x];
     p[x]:=len;
   end;

 procedure build(x,y,f:longint);
   begin
     add(x,y,f);
     add(y,x,);
   end;

 function sap:longint;
   var u,i,j,tmp,neck,q:longint;
   begin
     sap:=;
     u:=;
     fillchar(numh,sizeof(numh),);
     fillchar(h,sizeof(h),);
     for i:= to max do
       cur[i]:=p[i];
     cur[t]:=p[t];
     numh[]:=max+;
     neck:=inf;
     while h[]<max+ do
     begin
       d[u]:=neck;
       i:=cur[u];
       while i<>- do
       begin
         j:=e[i].po;
         if (e[i].flow>) and (h[u]=h[j]+) then
         begin
           neck:=min(neck,e[i].flow);
           pre[j]:=u;
           cur[u]:=i;
           u:=j;
           if u=t then
           begin
             sap:=sap+neck;
             while u<> do
             begin
               u:=pre[u];
               j:=cur[u];
               dec(e[j].flow,neck);
               inc(e[j xor ].flow,neck);
             end;
             neck:=inf;
           end;
           break;
         end;
         i:=e[i].next;
       end;
       if i=- then
       begin
         dec(numh[h[u]]);
         if numh[h[u]]= then break;
         q:=-;
         tmp:=max+;
         i:=p[u];
         while i<>- do
         begin
           j:=e[i].po;
           if (e[i].flow>) then
             if h[j]<tmp then
             begin
               tmp:=h[j];
               q:=i;
             end;
           i:=e[i].next;
         end;
         h[u]:=tmp+;
         cur[u]:=q;
         inc(numh[h[u]]);
         if u<> then
         begin
           u:=pre[u];
           neck:=d[u];
         end;
       end;
     end;
   end;

 begin
   len:=-;
   fillchar(p,sizeof(p),);
   readln(n,m,k);
   build(,,k);
   for i:= to m do
     readln(a[i],b[i],c[i]);
   t:=;
   ans:=;
   repeat
     for i:= to m do
       build(ans*n+a[i],(ans+)*n+b[i],c[i]);
     for i:= to n do
       build(ans*n+i,(ans+)*n+i,inf);
     build((ans+)*n+n,t,inf);
     max:=(ans+)*n+n;
     inc(ans);
     s:=s+sap;
   until s=k;
   writeln(ans);
 end.