好吧，是我智障啦。。。。。

我是P党，这是我在洛谷上发布的题解：

var dis:array[1..110,1..110] of double;//顾名思义，两点间的距离  
    x,y,tt:array[1..110] of longint;//用x y数组来记录各点的坐标,tt存储各城市公路的花费  
    n,xx,s,t,a,b,i,j,k:longint;  
    ans:double;  
  
procedure ce(t1,t2:longint);//计算两个点之间的花费  
  begin  
    dis[t1,t2]:=sqrt(sqr(x[t1]-x[t2])+sqr(y[t1]-y[t2]));  
    if ((t1-1) div 4)=((t2-1) div 4) then  
      dis[t1,t2]:=dis[t1,t2]*tt[((t1-1) div 4)+1]  
    else dis[t1,t2]:=dis[t1,t2]*t;  
    dis[t2,t1]:=dis[t1,t2];  
  end;  
  
function ver(t1,t2,t3:longint):longint;//找一个城市中已知三个点的顶点  
  begin  
    if (dis[t1,t2]>dis[t2,t3]) and (dis[t1,t2]>dis[t3,t1]) then ver:=t3;  
    if (dis[t2,t3]>dis[t1,t2]) and (dis[t2,t3]>dis[t3,t1]) then ver:=t1;  
    if (dis[t3,t1]>dis[t2,t3]) and (dis[t3,t1]>dis[t1,t2]) then ver:=t2;  
  end;  
  
procedure fpl(t1,t2,t3:longint);//计算第四个点的位置  t1,t2,t3为其他三个点的顺序  
  begin                  
    ce(t1,t2);  
    ce(t2,t3);  
    ce(t3,t1);  
      if ver(t1,t2,t3)=t1 then begin  
           x[t3+1]:=x[t3]+x[t2]-x[t1];  
           y[t3+1]:=y[t3]+y[t2]-y[t1];  
         end;  
      if ver(t1,t2,t3)=t2 then begin  
           x[t3+1]:=x[t3]+x[t1]-x[t2];  
           y[t3+1]:=y[t3]+y[t1]-y[t2];  
         end;  
      if ver(t1,t2,t3)=t3 then begin  
           x[t3+1]:=x[t1]+x[t2]-x[t3];  
           y[t3+1]:=y[t1]+y[t2]-y[t3];  
         end;  
    end;  
  
begin  
  readln(n);  
  xx:=0;  
  repeat  
    inc(xx);//计算已处理了多少组数据  
    ans:=maxlongint;  
    fillchar(dis,sizeof(dis),0);//初始化dis数组  
    readln(s,t,a,b);  
  
    for i:=1 to s do begin  
      readln(x[4*i-3],y[4*i-3],x[4*i-2],y[4*i-2],x[4*i-1],y[4*i-1],tt[i]);  
      fpl(4*i-3,4*i-2,4*i-1);  
    end;  
  
    for i:=1 to 4*s do //绘制地图  
      for j:=1 to 4*s do  
        ce(i,j);  
  
    for k:=1 to 4*s do //floyd处理  
      for i:=1 to 4*s do  
        for j:=1 to 4*s do  
          if dis[i,k]+dis[k,j]<dis[i,j] then  
            dis[i,j]:=dis[i,k]+dis[k,j];  
  
    for i:=4*a-3 to 4*a do //取最小值  
      for j:=4*b-3 to 4*b do  
        if dis[i,j]<ans then  
          ans:=dis[i,j];  
  
    writeln(ans:0:1);  
  until xx=n;  
end. 

数据不一样，程序鲁棒性不够，是不是这样？

我在Vijos上AC，结果到Luogu上WA40？？？？