CodeVS 1041 Car的旅行路线 Pascal

题目大意：

题解：很明显建图之后Floyd一下即可。

首先根据勾股定理，可以求出两点之间的距离为√(x1-x2)²(y1-y2)²，然后根据长方形内的三个点求出第四个点，也是根据勾股定理，先求出三个点之中每两个点之间的距离，即b1=√(x1-x2)²(y1-y2)²,b2=√(x1-x3)²(y1-y3)²,b3=√(x2-x3)²(y2-y3)²,如果最长边为b1，则第四个点的坐标为(x1+x2-x3,y1+y2-y3)；如果最长边为b2，则第四个点的坐标为(x1-x2+x3,y1-y2+y3)；如果最长边为b3，则第四个点的坐标为(-x1+x2+x3,-y1+y2+y3)。

把第四给点搞定之后，再在城市中四个点每两个点之间建一条边，权值为走公路所用花费。接下来在全部城市读入完毕之后，枚举每两个城市之间的每两个点，建一条边，权值为航班花费。

这样就建图完成了，接下来就进行Floyd。

最后枚举起点与终点的每两个点，求出最小花费即为答案。

var n,s,t,a,b,i,j,k,i1,j1,x4,y4,he,l:longint;
var ans:double;
var d:array[0..500,0..500] of double;
var x,y:array[0..100,1..4] of longint;
var b1:array[1..100] of longint;
function min(a,b:double):double;
begin
  if a<b then exit(a);exit(b);
end;
procedure swap(var a,b:longint);
var t:longint;
begin
  t:=a;a:=b;b:=t;
end;
procedure max(var a,b,c:longint);
var t:longint;
begin
  if a<c then swap(a,c);
  if a<b then swap(a,b);
  if b<c then swap(b,c);
end;
function hh(x,y,x1,y1:longint):double;
begin
  exit(sqrt(sqr(x-x1)+sqr(y-y1))*he);
end;
procedure xy(x1,y1,x2,y2,x3,y3:longint);
var b1,b2,b3:double;
begin
  he:=1;
  b1:=hh(x1,y1,x2,y2);b2:=hh(x1,y1,x3,y3);b3:=hh(x2,y2,x3,y3);
  if (b1>=b2) and (b1>=b3) then
  begin x[j,4]:=abs(x1+x2-x3);y[j,4]:=abs(y1+y2-y3);end
  else if (b2>=b1) and (b2>=b3) then
  begin x[j,4]:=abs(x1+x3-x2);y[j,4]:=abs(y1+y3-y2);end else 
  begin x[j,4]:=abs(x2+x3-x1);y[j,4]:=abs(y2+y3-y1);end;
end;
begin
  read(n);
  for i1:=1 to n do
  begin
    for j:=1 to 500 do for k:=1 to 500 do d[j,k]:=0;
    fillchar(b1,sizeof(b1),0);
    fillchar(x,sizeof(x),0);fillchar(y,sizeof(y),0);
    read(s,t,a,b);ans:=maxlongint;
    for j:=1 to s do
    begin
      for k:=1 to 3 do read(x[j,k],y[j,k]);read(b1[j]);
      xy(x[j,1],y[j,1],x[j,2],y[j,2],x[j,3],y[j,3]);
      he:=b1[j];
      for k:=3 downto 1 do
        for j1:=k-1 downto 0 do
        begin
          d[4*j-k,4*j-j1]:=hh(x[j,4-k],y[j,4-k],x[j,4-j1],y[j,4-j1]);
          d[4*j-j1,4*j-k]:=d[4*j-k,4*j-j1];
        end;
    end;
    he:=t;
    for j:=1 to s do
      for k:=j+1 to s do
        for l:=3 downto 0 do
          for j1:=3 downto 0 do
          begin
            d[4*j-l,4*k-j1]:=hh(x[j,4-l],y[j,4-l],x[k,4-j1],y[k,4-j1]);
            d[4*k-j1,4*j-l]:=d[4*j-l,4*k-j1];
          end;
    for k:=1 to 4*s do
      for i:=1 to 4*s do
        for j:=1 to 4*s do
          if d[i,k]+d[k,j]<d[i,j] then
          d[i,j]:=d[i,k]+d[k,j];
          for i:=3 downto 0 do
      for j:=3 downto 0 do
      begin
        ans:=min(d[a*4-i,b*4-j],ans);
      end;
    writeln(ans:0:1);
  end;
end.

　　感谢阅读！