poj 1860 Currency Exchange 解题报告

题目链接:http://poj.org/problem?id=1860

题目意思:给出 N 种 currency, M种兑换方式,Nick 拥有的的currency 编号S 以及他的具体的currency(V)。M 种兑换方式中每种用6个数描述: A, B, Rab, Cab, Rba, Cba。其中,Rab: 货币A 兑换 货币B 的汇率为Rab,佣金为Cab。Rba:货币B 兑换 货币 A 的汇率,佣金为Cba。假设含有的A货币是x,那么如果兑换成B,得到的货币B 就是:(x-Cab) * Rab。问从 货币S 经过一定次数的兑换,最终回归到货币S,能否使得 Nick 本来含有的 S 大。

思路我是借鉴这个人的:

http://blog.csdn.net/lyhvoyage/article/details/19281013

可以说,是用了Bellman_ford 的 逆向思维。传统的Bellman_ford 是用来求可以含有负边权的最短路径,且判断是否有负权回路。

这个题目希望我们验证是否存在正权回路:顶点的权值能不断增加,且能无限一直松弛下去。

不过初始化与传统的Bellman_ford 是不同的, dist[S] = V,其他dist[i] = 0 / 无穷小。当 S 到其他点的 距离能不断增大时,说明存在正权回路。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5
 6 const int maxn = 1e4 + 10;
 7 const int maxv = 100 + 5;
 8
 9 double has, dist[maxv];
10 int cnt, type;
11 int N, M;
12
13 struct node
14 {
15     int a, b;
16     double rate, commission;
17 }currency[maxn];
18
19 bool Bellman_ford()
20 {
21     for (int i = 1; i <= N; i++)
22         dist[i] = (i == type ? has : 0);
23     for (int i = 1; i < N; i++)
24     {
25         bool flag = false;
26         for (int j = 1; j < cnt; j++)
27         {
28             double t = (dist[currency[j].a] - currency[j].commission) * currency[j].rate;
29             if (t > dist[currency[j].b])
30             {
31                 dist[currency[j].b] = (dist[currency[j].a] - currency[j].commission) * currency[j].rate;
32                 flag = true;
33             }
34         }
35         if (!flag)
36             break;
37     }
38     for (int j = 1; j < cnt; j++)
39     {
40         double t = (dist[currency[j].a] - currency[j].commission) * currency[j].rate;
41         if (t > dist[currency[j].b])
42             return true;
43     }
44     return false;
45 }
46
47 int main()
48 {
49     while (scanf("%d%d%d%lf", &N, &M, &type, &has) != EOF)
50     {
51         int A, B;
52         double Rab, Cab, Rba, Cba;
53         cnt = 1;
54         while (M--)
55         {
56             scanf("%d%d%lf%lf%lf%lf", &A, &B, &Rab, &Cab, &Rba, &Cba);
57             currency[cnt].a = A;
58             currency[cnt].b = B;
59             currency[cnt].rate = Rab;
60             currency[cnt].commission = Cab;
61
62             currency[++cnt].a = B;
63             currency[cnt].b = A;
64             currency[cnt].rate = Rba;
65             currency[cnt++].commission = Cba;
66         }
67         printf("%s\n", Bellman_ford() ? "YES" : "NO");
68     }
69     return 0;
70 }

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时间: 2024-10-12 13:29:59

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