题解：
平面计算几何题。每次枚举相拼的两条边，在按顺序扫过两个多边形对应的变，检查是否是无缝屏接，拼好后是否是凸多边形。
注意给点的顺序不一定都是逆时针的，先按照给点的顺序扫过每一条边，记录下角度的偏转情况。若最终偏转为-2π，则为顺时针；若为2π，则为逆时针。
通过倒着排序给出的点，使两个多边形都变成是逆时针的，就可以正常做了。
代码：
var i,j,k,l,n,m:longint; a,b:array[-1000..2000,1..2]of extended; tt:array[1..2]of extended; e:extended; function dis(x,y,xx,yy:extended):extended; begin exit(sqrt(sqr(x-xx)+sqr(y-yy))); end; function xj2(x,y:extended):extended; var i,j,k,l:longint; begin if(abs(x)<0.0000001)and(y>0)then exit(pi/2); if(abs(x)<0.0000001)and(y<0)then exit(3*pi/2); if(abs(y)<0.0000001)and(x>0)then exit(0); if(abs(y)<0.0000001)and(x<0)then exit(pi); if(x>0)and(y>0)then exit(arctan(y/x)); if(x>0)and(y<0)then exit(2*pi-arctan(-y/x)); if(x<0)and(y>0)then exit(pi/2+arctan(-x/y)); if(x<0)and(y<0)then exit(pi+arctan(y/x)); end; function xj(x,y,xx,yy,xxx,yyy:extended):extended; var z1,z2:extended; begin z1:=xj2(x-xx,y-yy); z2:=xj2(xxx-xx,yyy-yy); while z2>0.0000001+z1 do z2:=z2-2*pi; exit(z1-z2); end; function ss:longint; var i,j,ii,jj,k,l:longint; z1,z2,z3:extended; begin for i:=1 to n do for j:=1 to m do if abs(dis(a[i-1,1],a[i-1,2],a[i,1],a[i,2]) -dis(b[j+1,1],b[j+1,2],b[j,1],b[j,2]))<0.0000001 then begin z1:=xj(a[i-2,1],a[i-2,2],a[i-1,1],a[i-1,2],a[i,1],a[i,2]); z2:=xj(b[j+2,1],b[j+2,2],b[j+1,1],b[j+1,2],b[j,1],b[j,2]); while z2>0.0000001+z1 do z2:=z2-2*pi; if z1-z2>pi+0.0000001 then continue; ii:=i+1; jj:=j-1; l:=0; while true do begin z1:=xj(a[ii-2,1],a[ii-2,2],a[ii-1,1],a[ii-1,2],a[ii,1],a[ii,2]); z2:=xj(b[jj+2,1],b[jj+2,2],b[jj+1,1],b[jj+1,2],b[jj,1],b[jj,2]); while z2>0.0000001+z1 do z2:=z2-2*pi; if z1-z2<0.0000001 then begin if abs(dis(a[ii-1,1],a[ii-1,2],a[ii,1],a[ii,2]) -dis(b[jj+1,1],b[jj+1,2],b[jj,1],b[jj,2]))<0.0000001 then begin inc(ii); dec(jj); end else begin l:=1; break; end; continue; end; if z1-z2<pi-0.0000001 then begin l:=1; break; end; break; end; if l=0 then begin for k:=ii to i+n-2 do if xj(a[k-1,1],a[k-1,2],a[k,1],a[k,2],a[k+1,1],a[k+1,2])>pi+0.0000001 then begin l:=1; break; end; for k:=jj downto j-m+2 do if xj(b[k+1,1],b[k+1,2],b[k,1],b[k,2],b[k-1,1],b[k-1,2])<pi-0.0000001 then begin l:=1; break; end; if l=0 then exit(1); end; end; exit(0); end; begin while not eof do begin readln(n); for i:=1 to n do readln(a[i,1],a[i,2]); a[n+1]:=a[1]; a[0]:=a[n]; e:=0; for i:=1 to n do e:=e+xj(a[i-1,1],a[i-1,2],a[i,1],a[i,2],a[i+1,1],a[i+1,2])-pi; if e>0 then begin for i:=1 to n div 2 do begin tt:=a[i]; a[i]:=a[n-i+1]; a[n-i+1]:=tt; end; end; for i:=1 to n do a[i-n]:=a[i]; for i:=1 to n do a[i+n]:=a[i]; readln(m); for i:=1 to m do readln(b[i,1],b[i,2]); b[m+1]:=b[1]; b[0]:=b[m]; e:=0; for i:=1 to m do e:=e+xj(b[i-1,1],b[i-1,2],b[i,1],b[i,2],b[i+1,1],b[i+1,2])-pi; if e>0 then begin for i:=1 to m div 2 do begin tt:=b[i]; b[i]:=b[m-i+1]; b[m-i+1]:=tt; end; end; for i:=1 to m do b[i-m]:=b[i]; for i:=1 to m do b[i+m]:=b[i]; writeln(ss); end; end.