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
1 const inf=1000007; 2 type node=record 3 point,next,flow:longint; 4 end; 5 6 var edge:array[0..800010] of node; 7 cur,p,h,numh,pre,d:array[0..20010] of longint; 8 x,y,z:array[0..200010] of longint; 9 i,len,u,v,l,ans,n,m:longint; 10 11 function min(a,b:longint):longint; 12 begin 13 if a>b then exit(b) else exit(a); 14 end; 15 16 procedure add(x,y,z:longint); 17 begin 18 inc(len); 19 edge[len].point:=y; 20 edge[len].flow:=z; 21 edge[len].next:=p[x]; 22 p[x]:=len; 23 end; 24 25 function sap(s,t:longint):longint; 26 var u,i,j,q,tmp,neck:longint; 27 begin 28 for i:=1 to n do 29 cur[i]:=i; 30 fillchar(numh,sizeof(numh),0); 31 fillchar(h,sizeof(h),0); 32 u:=s; 33 h[s]:=0; 34 numh[0]:=n; 35 neck:=inf; 36 sap:=0; 37 while h[s]<n do 38 begin 39 d[u]:=neck; 40 i:=cur[u]; 41 while i<>-1 do 42 begin 43 j:=edge[i].point; 44 if (edge[i].flow>0) and (h[u]=h[j]+1) then 45 begin 46 cur[u]:=i; 47 pre[j]:=u; 48 neck:=min(neck,edge[i].flow); 49 u:=j; 50 if u=t then 51 begin 52 sap:=sap+neck; 53 while u<>s do 54 begin 55 u:=pre[u]; 56 j:=cur[u]; 57 dec(edge[j].flow,neck); 58 inc(edge[j xor 1].flow,neck); 59 end; 60 neck:=inf; 61 end; 62 break; 63 end; 64 i:=edge[i].next; 65 end; 66 if i=-1 then 67 begin 68 dec(numh[h[u]]); 69 if numh[h[u]]=0 then exit; 70 q:=-1; 71 tmp:=n; 72 i:=p[u]; 73 while i<>-1 do 74 begin 75 j:=edge[i].point; 76 if (edge[i].flow>0) then 77 if h[j]<tmp then 78 begin 79 tmp:=h[j]; 80 q:=i; 81 end; 82 i:=edge[i].next; 83 end; 84 cur[u]:=q; 85 h[u]:=tmp+1; 86 inc(numh[h[u]]); 87 if u<>s then 88 begin 89 u:=pre[u]; 90 neck:=d[u]; 91 end; 92 end; 93 end; 94 end; 95 96 begin 97 readln(n,m); 98 for i:=1 to m do 99 readln(x[i],y[i],z[i]); 100 readln(u,v,l); 101 fillchar(p,sizeof(p),255); 102 len:=-1; 103 for i:=1 to m do 104 if z[i]>l then 105 begin 106 add(x[i],y[i],1); 107 add(y[i],x[i],0); 108 add(y[i],x[i],1); 109 add(x[i],y[i],0); 110 end; 111 ans:=sap(u,v); 112 113 fillchar(p,sizeof(p),255); 114 len:=-1; 115 for i:=1 to m do 116 if z[i]<l then 117 begin 118 add(x[i],y[i],1); 119 add(y[i],x[i],0); 120 add(y[i],x[i],1); 121 add(x[i],y[i],0); 122 end; 123 ans:=ans+sap(u,v); 124 writeln(ans); 125 end.