bzoj2561

对于新加入的边，必须要既可能在最小生成树上也可能在最大生成树上
我们先对于最小生成树考虑
根据kruskal的理论，不难发现，u--v 长度为L的边可能出现在最小生成树上
就是说删边剩下的比L小的边一定不能使u，v连通，
因此问题就转化为求u,v两点的最小割了
最大生成树同理，最后答案是两个加起来

const inf=1000007;
type node=record
       point,next,flow:longint;
     end;

var edge:array[0..800010] of node;
    cur,p,h,numh,pre,d:array[0..20010] of longint;
    x,y,z:array[0..200010] of longint;
    i,len,u,v,l,ans,n,m:longint;

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

procedure add(x,y,z:longint);
  begin
    inc(len);
    edge[len].point:=y;
    edge[len].flow:=z;
    edge[len].next:=p[x];
    p[x]:=len;
  end;

function sap(s,t:longint):longint;
  var u,i,j,q,tmp,neck:longint;
  begin
    for i:=1 to n do
      cur[i]:=i;
    fillchar(numh,sizeof(numh),0);
    fillchar(h,sizeof(h),0);
    u:=s;
    h[s]:=0;
    numh[0]:=n;
    neck:=inf;
    sap:=0;
    while h[s]<n do
    begin
      d[u]:=neck;
      i:=cur[u];
      while i<>-1 do
      begin
        j:=edge[i].point;
        if (edge[i].flow>0) and (h[u]=h[j]+1) then
        begin
          cur[u]:=i;
          pre[j]:=u;
          neck:=min(neck,edge[i].flow);
          u:=j;
          if u=t then
          begin
            sap:=sap+neck;
            while u<>s do
            begin
              u:=pre[u];
              j:=cur[u];
              dec(edge[j].flow,neck);
              inc(edge[j xor 1].flow,neck);
            end;
            neck:=inf;
          end;
          break;
        end;
        i:=edge[i].next;
      end;
      if i=-1 then
      begin
        dec(numh[h[u]]);
        if numh[h[u]]=0 then exit;
        q:=-1;
        tmp:=n;
        i:=p[u];
        while i<>-1 do
        begin
          j:=edge[i].point;
          if (edge[i].flow>0) then
            if h[j]<tmp then
            begin
              tmp:=h[j];
              q:=i;
            end;
          i:=edge[i].next;
        end;
        cur[u]:=q;
        h[u]:=tmp+1;
        inc(numh[h[u]]);
        if u<>s then
        begin
          u:=pre[u];
          neck:=d[u];
        end;
      end;
    end;
  end;

begin
  readln(n,m);
  for i:=1 to m do
    readln(x[i],y[i],z[i]);
  readln(u,v,l);
  fillchar(p,sizeof(p),255);
  len:=-1;
  for i:=1 to m do
    if z[i]>l then
    begin
      add(x[i],y[i],1);
      add(y[i],x[i],0);
      add(y[i],x[i],1);
      add(x[i],y[i],0);
    end;
  ans:=sap(u,v);

  fillchar(p,sizeof(p),255);
  len:=-1;
  for i:=1 to m do
    if z[i]<l then
    begin
      add(x[i],y[i],1);
      add(y[i],x[i],0);
      add(y[i],x[i],1);
      add(x[i],y[i],0);
    end;
  ans:=ans+sap(u,v);
  writeln(ans);
end.