PKU 3311 Hie with the Pie 最短路 folyd

题意/Description：

    The Pizazz Pizzeria prides itself in delivering pizzas to its customers as fast as possible. Unfortunately, due to cutbacks, they can afford to hire only one driver to do the deliveries. He will wait for 1 or more (up to 10) orders to be processed before he starts any deliveries. Needless to say, he would like to take the shortest route in delivering these goodies and returning to the pizzeria, even if it means passing the same location(s) or the pizzeria more than once on the way. He has commissioned you to write a program to help him.

 

读入/Input：

    Input will consist of multiple test cases. The first line will contain a single integer n indicating the number of orders to deliver, where 1 ≤ n ≤ 10. After this will be n + 1 lines each containing n + 1 integers indicating the times to travel between the pizzeria (numbered 0) and the n locations (numbers 1 to n). The jth value on the ith line indicates the time to go directly from location i to location j without visiting any other locations along the way. Note that there may be quicker ways to go from i to jvia other locations, due to different speed limits, traffic lights, etc. Also, the time values may not be symmetric, i.e., the time to go directly from location i to j may not be the same as the time to go directly from location j to i. An input value of n = 0 will terminate input.

 

输出/Output：

    For each test case, you should output a single number indicating the minimum time to deliver all of the pizzas and return to the pizzeria.

 

题解/solution：

 这题听一个姓刘的说，要用folyd+状态压缩DP，听起来好恐怖。给出DP式子：

    f[k,i]:=f[k xor (1 shl (i-1)),j]+a[j,i]; (k xor (1 shl (i-1))<>0)，((1 shl (j-1))<>0)

  乍一看真的好恐怖。

  解释一下：DP式子指：在没经过城市I的状态中，寻找合适的中间点J使得距离更短，和FLOYD一样。

           条件是指：状态S中已经过城市i  ，枚举不是城市I的其他城市。

  然后自己敲吧。

 

代码/Code：

(部分程序)

procedure main;
var
  j,i,k:longint;
begin
  fillchar(f,sizeof(f),$7f);
  f[0,1]:=0;
  for k:=0 to 1 shl n-1 do
    begin
      for i:=1 to n do
        begin
          if k=1 shl (i-1) then f[k,i]:=a[0,i];
          if (k and (1 shl (i-1)))<>0 then
            begin
              for j:=1 to n do
                if (k xor (1 shl (i-1))<>0) and ((1 shl (j-1))<>0) then
                  if f[k xor (1 shl (i-1)),j]+a[j,i]<f[k,i] then
                    f[k,i]:=f[k xor (1 shl (i-1)),j]+a[j,i];
            end;
        end;
    end;
end;

procedure print;
var
  i:longint;
begin
  ans:=maxlongint;
  for i:=1 to n do
    if ans>f[1 shl n-1,i]+a[i,0] then ans:=f[1 shl n-1,i]+a[i,0];
  writeln(ans);
end;