[POJ1789 Truck History]

[题目来源]：CTU Open 2003

[关键字]：图论

[题目大意]：给出一个n*7的字母矩阵，每一行代表一种车，那么定义的每种车之间的距离为七个位置上每个相应的位置上不同字母数的和，要求的是不同种车的距离的和得最小值。

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[分析]：其实就是最小生成树的算法，由于是稠密图所以prim效率更高。

[代码]：

View Code

 1 {
 2 PROB:POJ1789
 3 DATE:2011\10\14
 4 }
 5 program project1;
 6 var
 7   n: longint;
 8   w: array[0..2010] of string[7];
 9   map: array[0..2010,0..2010] of byte;
10   d: array[0..2010] of longint;
11   b: array[0..2011] of boolean;
12 
13 function dis(s1, s2: string):longint;
14 var
15   i, sum: longint;
16 begin
17   sum := 0;
18   for i := 1 to 7 do
19     if s1[i] <> s2[i] then inc(sum);
20   exit(sum);
21 end;
22 
23 procedure init;
24 var
25   i, j: longint;
26 begin
27   readln(n);
28   if n = 0 then halt;
29   for i := 1 to n do
30     readln(w[i]);
31   for i := 1 to n do
32     for j := i+1 to n do
33       begin
34         map[i,j] := dis(w[i],w[j]);
35         map[j,i] := map[i,j];
36       end;
37 end;
38 
39 procedure prim;
40 var
41   i, j, p, ans, min: longint;
42 begin
43   fillchar(b,sizeof(b),false);
44   fillchar(d,sizeof(d),100);
45   d[1] := 0;
46   ans := 0;
47   for i := 1 to n do
48     begin
49       min := maxlongint;
50       for j := 1 to n do
51         if (not b[j]) and (min > d[j]) then
52           begin
53             min := d[j];
54             p := j;
55           end;
56       inc(ans,min);
57       b[p] := true;
58       for j := 1 to n do
59         if (not b[j]) and (d[j] > map[p,j]) then
60           d[j] := map[p,j];
61     end;
62   writeln('The highest possible quality is 1/',ans,'.');
63 end;
64 
65 begin
66   while 1 = 1 do
67     begin
68       init;
69       prim;
70     end;
71 end.