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[0..200,0..200] of boolean;
    f:array[0..200,0..200] of int64;
    g:array[0..200] of int64;
    q,rd:array[0..200] of longint;
    a,b,c,d,n,m,x,y,i,j,k,h,t:longint;
    ans:int64;

begin
  readln(n,m);
  for i:=1 to m do
  begin
    readln(x,y);
    w[x,y]:=true;
    inc(rd[y]);
  end;
  for i:=1 to n do
    if rd[i]=0 then
    begin
      inc(t);
      q[t]:=i;
    end;
  h:=1;
  while h<=t do
  begin
    x:=q[h];
    for i:=1 to n do
      if w[x,i] then
      begin
        dec(rd[i]);
        if rd[i]=0 then
        begin
          inc(t);
          q[t]:=i;
        end;
      end;

    inc(h);
  end;
  readln(a,b,c,d);
  for i:=1 to n do
    for j:=i to n do
      if q[i]=q[j] then f[q[i],q[j]]:=1
      else begin
        for k:=i to j-1 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:=1 to n do
  begin
    if (f[q[i],b]<>0) and (f[q[i],d]<>0) then
    begin
      g[q[i]]:=f[a,q[i]]*f[c,q[i]];
      for j:=1 to i-1 do
        g[q[i]]:=g[q[i]]-g[q[j]]*f[q[j],q[i]]*f[q[j],q[i]];
    end;
  end;
  for i:=1 to n do
    ans:=ans-g[i]*f[i,b]*f[i,d];
  ans:=ans+f[a,b]*f[c,d];
  writeln(ans);
end.