POJ 1860 Currency Exchange

题目大意:假設有A，B兩種貨幣，要將A換成B，須透過匯率Rab和手續費Cab，因此實際得到B貨幣=(A-Cab)*Rab元。

第一行輸入N, M, S, V，N表示共有N種貨幣(1<=N<=100)，M表示底下有M行，每一行有六個數字A,B,Rab,Cab,Rba,Cab，Rab表示A換成B的匯率，Cab表示A換成B需要扣除的手續費，Rba和Cba同理。現在Nick有第S種貨幣V元，題目問Nick透過不斷交換貨幣，然後最後換回第S種貨幣的時候能否大於V元?

解题思路:Bellman-ford改一改，只要能够无限松弛，就符合题意（我这里优化了一下Bellman-ford，就是不能松弛了，就退出松弛循环（这道题可以直接判断不行了）， 感觉有点像冒泡的优化

 1 #include <iostream>
 2 #include <cstring>
 3 #include <algorithm>
 4 #define MAXVERTEXNUM 500
 5 #define INF 1000000
 6 using namespace std;
 7 
 8 struct E
 9 {
10     int a, b;
11     double exchange_currency, tips;
12 }Edge[MAXVERTEXNUM];
13 int Nv, Ne, s, num = 0;
14 double org, dist[MAXVERTEXNUM];
15 
16 bool Bellman_Ford()
17 {
18     for (int i = 1; i < Nv; ++i)
19     {
20         bool flag = false;
21         for (int j = 0; j < num; ++j)
22         {
23             if (dist[Edge[j].b] < (dist[Edge[j].a] - Edge[j].tips) * Edge[j].exchange_currency)
24             {
25                 dist[Edge[j].b] = (dist[Edge[j].a] - Edge[j].tips) * Edge[j].exchange_currency;
26                 flag = true;
27             }
28         }
29         if (!flag)
30             return false;//写break还要快一些，黑人问号？？？
31     }
32 
33     for (int i = 0; i < num; ++i)
34         if (dist[Edge[i].b] < (dist[Edge[i].a] - Edge[i].tips) * Edge[i].exchange_currency)
35             return true;
36 
37     return false;
38 }
39 
40 int main()
41 {
42     ios::sync_with_stdio(false);
43 
44     cin >> Nv >> Ne >> s >> org;
45     memset(dist, 0, sizeof(dist));
46     dist[s] = org;
47     //init
48 
49     for (int i = 1; i <= Ne; ++i)
50     {
51         int V1, V2;
52         double t1, t2, t3, t4;
53         cin >> V1 >> V2 >> t1 >> t2 >> t3 >> t4;
54         Edge[num].a = V1, Edge[num].b = V2;
55         Edge[num].exchange_currency = t1;
56         Edge[num++].tips = t2;
57         Edge[num].a = V2, Edge[num].b = V1;
58         Edge[num].exchange_currency = t3;
59         Edge[num++].tips = t4;
60     }
61 
62     if (Bellman_Ford())
63         cout << "YES" << endl;
64     else
65         cout << "NO" << endl;
66 
67     return 0;
68 }

 

posted @ 2018-06-26 22:31  duck_lu  阅读(139)  评论(0)  编辑  收藏  举报