【NOIP2014八校联考第3场第1试10.4】规避(path) (Standard IO)

题意：

求一条路径，不能经过规定的危险区域。

思路：

超级暴力枚举，每次判断两点之间是否可以连线，用计算几何算是否有交点。

程序：

没ac，70分的

 type    Point=record        x,y:real;    end;const maxn=300;var a:array [1..maxn] of longint; f:array [1..maxn,1..2] of longint; s:array [1..maxn,1..maxn] of real; i,j,n,x,y,p,k:longint; e,b,c,d:Point;function max(x,y:real):real;begin if x>y then exit(x)        else exit(y);end;function min(x,y:real):real;begin if x>y then exit(y)        else exit(x);end;function mult(a,b,c:Point):double;begin    exit((a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y));end;function kua(aa,bb,cc,dd:Point):boolean;begin    if (max(aa.x, bb.x) <= min(cc.x, dd.x)) then    begin        exit(false);    end;    if (max(aa.y, bb.y) <= min(cc.y, dd.y)) then    begin        exit(false);    end;    if (max(cc.x, dd.x) <= min(aa.x, bb.x)) then    begin        exit(false);    end;    if (max(cc.y, dd.y) <= min(aa.y, bb.y)) then    begin        exit(false);    end;    if (mult(cc, bb, aa)*mult(bb, dd, aa)<=0) then    begin        exit(false);    end;    if (mult(aa, dd, cc)*mult(dd, bb, cc)<=0) then    begin        exit(false);    end;    exit(true);end;function cheak(x1,y1,x2,y2:longint):boolean;var i,j,k,sum:longint;begin sum:=0; for i:=1 to n do begin  for j:=1 to a[i] do   for k:=j+2 to a[i] do   begin    e.x:=x1; e.y:=y1;    b.x:=x2; b.y:=y2;    c.x:=f[sum+j,1]; c.y:=f[sum+j,2];    d.x:=f[sum+k,1]; d.y:=f[sum+k,2];    if kua(e,b,c,d) then exit(false);   end;  sum:=sum+a[i]; end; exit(true);end;begin readln(n); for i:=1 to maxn do  for j:=1 to maxn do   s[i,j]:=999999999; for i:=1 to n do begin  readln(a[i]);  for j:=1 to a[i] do  begin   inc(p);   readln(f[p,1],f[p,2]);  end; end; for i:=1 to p do  for j:=i+1 to p  do   if cheak(f[i,1],f[i,2],f[j,1],f[j,2]) then   begin    s[i,j]:=sqrt(sqr(f[i,1]-f[j,1])+sqr(f[i,2]-f[j,2]));    s[j,i]:=s[i,j];   end; for i:=1 to p do  for j:=1 to p do   if i<>j then   for k:=1 to p do   if s[i,j]>s[i,k]+s[k,j] then    s[i,j]:=s[i,k]+s[k,j]; readln(x,y); writeln(s[x,y]:0:4)end.