[POJ1860 Currency Exchange]

[题目来源]：Northeastern Europe 2001, Northern Subregion

[关键字]：图论

[题目大意]：套汇问题,汇率为rab,增加一个手续费cab,那么每次套汇的结果是(本金-手续费cab)*汇率rab,现在给出一个人所拥有的货币类型以及拥有该类型货币的总量,需要判断的是经过一系列货币交换后,他能不能是他的货币增值.

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[分析]：其实就是构建出差分约束系统求图中是否存在环。唯一要注意的是由于松弛条件不是一般的加或减，所以松弛不能只作n-1次，而要一直松弛到出现环(d[st]变大)，或不能在松弛。

[代码]：

View Code

 1 program Project1;
 2 type
 3   rec = record
 4     x, y: longint;
 5     cxy, rxy: real;
 6   end;
 7 var
 8   n, m, s, tot: longint;
 9   ds: real;
10   d: array[0..1000] of real;
11   e: array[0..2000] of rec;
12 
13 procedure make(x, y: longint; rxy, cxy: real);
14 begin
15   inc(tot);
16   e[tot].x := x;
17   e[tot].y := y;
18   e[tot].cxy := cxy;
19   e[tot].rxy := rxy;
20 end;
21 
22 procedure init;
23 var
24   i, x, y: longint;
25   rxy, cxy, ryx, cyx: real;
26 begin
27   readln(n,m,s,ds);
28   for i := 1 to n do d[i] := -9999999;
29   d[s] := ds;
30   for i := 1 to m do
31     begin
32       readln(x,y,rxy,cxy,ryx,cyx);
33       make(x,y,rxy,cxy);
34       make(y,x,ryx,cyx);
35     end;
36 end;
37 
38 procedure print(x: longint);
39 begin
40   if x = 1 then writeln('YES');
41   if x = 0 then writeln('NO');
42   //readln;
43   //readln;
44   halt;
45 end;
46 
47 procedure bellman;
48 var
49   i, j: longint;
50   f: boolean;
51 begin
52   while 1 = 1 do
53     begin
54       f := true;
55       for j := 1 to tot do
56         if d[e[j].y] < (d[e[j].x]-e[j].cxy)*e[j].rxy then
57           begin
58             d[e[j].y] := (d[e[j].x]-e[j].cxy)*e[j].rxy;
59             f := false;
60           end;
61        if d[s] > ds then print(1);
62        if f then print(0);
63     end;
64 end;
65 
66 begin
67   init;
68   bellman
69 end.