poj 1860 Currency Exchange 解题报告

 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 }