bzoj1471

转化补集的思想，首先求出任意两点之间路径数目

然后求两条路径第一次相交在点k（按照拓扑排序的顺序）的数目，显然这里要用到容斥

然后pascal有坑爹的范围检测，所以运算中有些不会影响到答案但会爆int64的地方要判断一下

 const inf=1e18;
 var w:array[..,..] of boolean;
     f:array[..,..] of int64;
     g:array[..] of int64;
     q,rd:array[..] of longint;
     a,b,c,d,n,m,x,y,i,j,k,h,t:longint;
     ans:int64;

 begin
   readln(n,m);
   for i:= to m do
   begin
     readln(x,y);
     w[x,y]:=true;
     inc(rd[y]);
   end;
   for i:= to n do
     if rd[i]= then
     begin
       inc(t);
       q[t]:=i;
     end;
   h:=;
   while h<=t do
   begin
     x:=q[h];
     for i:= to n do
       if w[x,i] then
       begin
         dec(rd[i]);
         if rd[i]= then
         begin
           inc(t);
           q[t]:=i;
         end;
       end;

     inc(h);
   end;
   readln(a,b,c,d);
   for i:= to n do
     for j:=i to n do
       if q[i]=q[j] then f[q[i],q[j]]:=
       else begin
         for k:=i to j- do
           if w[q[k],q[j]] then
             if f[q[i],q[j]]+f[q[i],q[k]]<inf then f[q[i],q[j]]:=f[q[i],q[j]]+f[q[i],q[k]];
       end;

   for i:= to n do
   begin
     if (f[q[i],b]<>) and (f[q[i],d]<>) then
     begin
       g[q[i]]:=f[a,q[i]]*f[c,q[i]];
       for j:= to i- do
         g[q[i]]:=g[q[i]]-g[q[j]]*f[q[j],q[i]]*f[q[j],q[i]];
     end;
   end;
   for i:= to n do
     ans:=ans-g[i]*f[i,b]*f[i,d];
   ans:=ans+f[a,b]*f[c,d];
   writeln(ans);
 end.