[POJ2240 Arbitrage]

[题目来源]：POJ2240

[关键字]：判断环

[题目大意]：给出一些汇率，问是否能将手中的钱通过兑换，使最后再换回本币时数量增加。

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[分析]：就是判断给出的图中是否存在一条正权环。但是要注意的是，因为路径权值不再是单纯相加，而是有了乘法就不能保证n-1次松弛后就能判断环，可以将结束条件改为：1、不能再松弛（无环）。2、源点已增大（有环）。

[代码]：

View Code

  1 program Project1;
  2 type
  3   rec = record
  4     x, y: longint;
  5     d: real;
  6   end;
  7 var
  8   n, m, tc: longint;
  9   w: array[0..100] of string;
 10   d: array[0..200] of real;
 11   e: array[0..2000] of rec;
 12   
 13 function findname(s: string):longint;
 14 var
 15   i: longint;
 16 begin
 17   for i := 1 to n do
 18     if w[i] = s then exit(i);
 19 end;
 20 
 21 procedure make(i, x, y: longint; d: real);
 22 begin
 23   e[i].x := x;
 24   e[i].y := y;
 25   e[i].d := d;
 26 end;
 27 
 28 procedure init;
 29 var
 30   i, x, y: longint;
 31   d: real;
 32   s: string;
 33   ch: char;
 34 begin
 35   for i := 1 to n do
 36     readln(w[i]);
 37   readln(m);
 38   for i := 1 to m do
 39     begin
 40       read(ch);
 41       s := '';
 42       while ch <> ' ' do
 43         begin
 44           s := s+ch;
 45           read(ch);
 46         end;
 47       x := findname(s);
 48       read(d);
 49       read(ch);
 50       readln(s);
 51       y := findname(s);
 52       make(i,x,y,d);
 53     end;
 54   readln;
 55   //for i := 1 to m do writeln(e[i].x,' ',e[i]. y,' ',e[i].d);
 56 end;
 57 
 58 function bellman(st: longint):boolean;
 59 var
 60   i: longint;
 61   f: boolean;
 62 begin
 63   for i := 1 to n do d[i] := -maxlongint;
 64   d[st] := 1;
 65   while 1 = 1 do
 66     begin
 67       f := true;
 68       for i := 1 to m do
 69         if d[e[i].y] < d[e[i].x]*e[i].d then
 70           begin
 71             d[e[i].y] := d[e[i].x]*e[i].d;
 72             f := false;
 73           end;
 74       if d[st] > 1 then exit(true);
 75       if f then exit(false);
 76     end;
 77 end;
 78 
 79 procedure work;
 80 var
 81   i: longint;
 82 begin
 83   for i := 1 to n do
 84     if bellman(i) then
 85       begin
 86         writeln('Case ',tc,': Yes');
 87         exit;
 88       end;
 89   writeln('Case ',tc,': No');
 90 end;
 91 
 92 begin
 93   readln(n);
 94   tc := 0;
 95   while n <> 0 do
 96     begin
 97       inc(tc);
 98       init;
 99       work;
100       readln(n);
101     end;
102 end.