codevs1041

题目地址：http://codevs.cn/problem/1041/

分析：

首先求出第四个点坐标

先根据的读入的三个点确定哪个是直角顶点，设为A B C 三点其中A为直角顶点

然后设另一点为D，连BC取中点E，则AD的中点也为E再由线段中点坐标公式((x1+x2)/2,(y1+y2)/2)可推知x1+x4=x2+x3 y1+y4=y2+y3 这就求出了第四点

将每个飞机场看成一个节点，则有4*s个节点

构建两个虚节点，一个作为0节点，另一个作为4*s+1，然后求出任两点的距离，然后就赤裸裸的最短路了
代码：

program travel;

var i,j,k,l,t1,s,c,d,n,c12,c23,c13,j1,j2,k1,k2:longint;

min1:real;

t:array[1..100]of longint;

f:array[0..401,0..401]of real;

cost:array[0..401]of real;

b:array[0..401]of boolean;

a:array[1..100,1..4,1..2]of longint;

procedure init;

begin

readln(s,t1,c,d);

for j:=1 to s do

begin

for k:=1 to 3 do

for l:=1 to 2 do

read(a[j,k,l]);

readln(t[j]);

end;

end;

procedure swap(var a,b:longint);

var x:longint;

begin

x:=a;a:=b;b:=x;

end;

function dis(var x1,y1,x2,y2:longint):real;

begin

exit(sqrt(sqr(x1-x2)+sqr(y1-y2)));

end;

procedure fourth;

begin

for j:=1 to s do

begin

c12:=sqr(a[j,1,1]-a[j,2,1])+sqr(a[j,1,2]-a[j,2,2]);

c13:=sqr(a[j,1,1]-a[j,3,1])+sqr(a[j,1,2]-a[j,3,2]);

c23:=sqr(a[j,2,1]-a[j,3,1])+sqr(a[j,2,2]-a[j,3,2]);

if c12+c23=c13 then begin swap(a[j,1,1],a[j,2,1]);swap(a[j,1,2],a[j,2,2]);end

else if c13+c23=c12 then begin swap(a[j,1,1],a[j,3,1]);swap(a[j,1,2],a[j,3,2]);end;

a[j,4,1]:=a[j,2,1]+a[j,3,1]-a[j,1,1];a[j,4,2]:=a[j,2,2]+a[j,3,2]-a[j,1,2];

end;

end;

function min(a,b:real):real;

begin

if a<b then exit(a) else exit(b);

end;

procedure count;

begin

for j:=1 to 4*s do

for k:=1 to 4*s do

if j<>k then

begin

j1:=(j-1)div 4+1;

j2:=(j-1)mod 4+1;

k1:=(k-1)div 4+1;

k2:=(k-1)mod 4+1;

if j1=k1 then

begin

f[j,k]:=dis(a[j1,j2,1],a[j1,j2,2],a[k1,k2,1],a[k1,k2,2])*t[j1];

if f[j,k]>dis(a[j1,j2,1],a[j1,j2,2],a[k1,k2,1],a[k1,k2,2])*t1 then

f[j,k]:=dis(a[j1,j2,1],a[j1,j2,2],a[k1,k2,1],a[k1,k2,2])*t1;

f[k,j]:=f[j,k];

end

else

begin

f[j,k]:=dis(a[j1,j2,1],a[j1,j2,2],a[k1,k2,1],a[k1,k2,2])*t1;

f[k,j]:=f[j,k];

end;

end;

end;

procedure dijkstra;

begin

for j:=1 to 4*s+1 do f[0,j]:=maxlongint;

for j:=1 to 4 do

f[0,c*4-4+j]:=0;

for j:=1 to 4*s do f[j,4*s+1]:=maxlongint;

for j:=1 to 4 do

f[d*4-4+j,4*s+1]:=0;

for j:=1 to 4*s+1 do

cost[j]:=f[0,j];

for j:=1 to 4*s+1 do

begin

min1:=maxlongint;l:=-1;

for k:=1 to 4*s+1 do

if(not b[k])and(cost[k]<min1) then

begin

min1:=cost[k];

l:=k;

end;

if l=-1 then break;

b[l]:=true;

for k:=0 to 4*s+1 do

if cost[l]+f[l,k]<cost[k] then

cost[k]:=cost[l]+f[l,k];

end;

end;

procedure main;

begin

readln(n);

for i:=1 to n do

begin

init;

fourth;

count;

dijkstra;

writeln(cost[4*s+1]:0:1);

end;

end;

begin

main;

end.