【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.

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