[NOIP 2001提高组T4]Car的旅行路线

[题目来源]：NOIP2001提高组T4

[关键字]：最短路径

[题目大意]：给定平面直角若干个矩形，计算（可经过其他矩形）两个矩形任意顶点间的最短路程费用。

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[分析]：其实题目本事没有太大的难点，只需要对每两个点进行连边（其实不用连知道坐标后现求两点间距离）然后求最短路即可。关键是如何知道给定三个顶点的矩形的另一个顶点。

公式：(x1,y1)(x2,y2)(x3,y3)为三个顶点坐标，如果(x1-x2)*(x3-x1)+(y1-y2)*(y3-y1)=0则x4=x2+x3-x1 y4=y2+y3-y1。只要循环直到找到第四个点即可，具体实现可看代码。 

[代码]：

View Code

  1 program project1;
  2 type
  3   rec = record
  4     x, y, c: longint;
  5   end;
  6 var
  7   n, tot, costf, st, ed, tc: longint;
  8   pos: array[0..2000] of rec;
  9   x, y: array[1..4] of longint;
 10   costt: array[0..200] of longint;
 11   b: array[0..200] of boolean;
 12   d: array[0..100] of real;
 13   
 14 procedure init;
 15 var
 16   i, j, k: longint;
 17 begin
 18   readln(n,costf,st,ed);
 19   for i := 1 to n do
 20     begin
 21       readln(x[1],y[1],x[2],y[2],x[3],y[3],costt[i]);
 22       for j := 1 to 3 do
 23         for k := 1 to 3 do
 24           if j <> k then
 25             if (x[j]-x[k])*(x[6-k-j]-x[j])+(y[j]-y[k])*(y[6-k-j]-y[j]) = 0 then
 26               begin
 27                 x[4] := x[k]+x[6-k-j]-x[j];
 28                 y[4] := y[k]+y[6-k-j]-y[j];
 29               end;
 30       for j := 1 to 4 do
 31         begin
 32           inc(tot);
 33           pos[tot].x := x[j];
 34           pos[tot].y := y[j];
 35           pos[tot].c := i;
 36         end;
 37     end;
 38   //for i := 1 to tot do writeln(pos[i].x,' ',pos[i].y,' ',pos[i].c);
 39 end;
 40 
 41 function dis(x, y: longint):real;
 42 begin
 43   dis := sqrt(sqr(pos[x].x-pos[y].x)+sqr(pos[x].y-pos[y].y));
 44   if pos[x].c = pos[y].c then
 45     dis := dis*costt[pos[x].c]
 46   else dis := dis*costf;
 47 end;
 48 
 49 function dij(st: longint):real;
 50 var
 51   i, j, p: longint;
 52   min: real;
 53 begin
 54   fillchar(b,sizeof(b),false);
 55   fillchar(d,sizeof(d),100);
 56   d[st] := 0;
 57   for i := 1 to tot do
 58     begin
 59       min := 1e38;
 60       for j := 1 to tot do
 61         if (not b[j]) and (d[j] < min) then
 62           begin
 63             min := d[j];
 64             p := j;
 65           end;
 66       b[p] := true;
 67       for j := 1 to tot do
 68         if not b[j] then
 69           if d[j] > d[p]+dis(p,j) then d[j] := d[p]+dis(p,j);
 70     end;
 71   dij := 1e38;
 72   for i := 1 to tot do
 73     begin
 74       if pos[i].c = ed then
 75         for j := 0 to 3 do
 76           if d[i+j] < dij then dij := d[i+j];
 77       if pos[i].c = ed then break;
 78     end;
 79 end;
 80 
 81 procedure work;
 82 var
 83   i, j: longint;
 84   min, temp: real;
 85 begin
 86   min := 1e38;
 87   for i := 1 to tot do
 88     begin
 89       if pos[i].c = st then
 90         for j := 0 to 3 do
 91           begin
 92             temp := dij(i+j);
 93             if temp < min then min := temp;
 94           end;
 95       if pos[i].c = st then break;
 96     end;
 97   writeln(min:0:1);
 98 end;
 99 
100 begin
101   assign(input,'d:\in.txt');reset(input);
102   assign(output,'d:\out.txt');rewrite(output);
103   readln(tc);
104   while tc > 0 do
105     begin
106       init;
107       work;
108       dec(tc);
109     end;
110   close(input);
111   close(output);
112 end.