HDU1217--Arbitrage

 

Arbitrage
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3157 Accepted Submission(s): 1431


Problem Description
Arbitrage is the use of discrepancies in currency exchange rates to transform one unit of a currency into more than one unit of the same currency. For example, suppose that 1 US Dollar buys 0.5 British pound, 1 British pound buys 10.0 French francs, and 1 French franc buys 0.21 US dollar. Then, by converting currencies, a clever trader can start with 1 US dollar and buy 0.5 * 10.0 * 0.21 = 1.05 US dollars, making a profit of 5 percent.

Your job is to write a program that takes a list of currency exchange rates as input and then determines whether arbitrage is possible or not.


Input
The input file will contain one or more test cases. Om the first line of each test case there is an integer n (1<=n<=30), representing the number of different currencies. The next n lines each contain the name of one currency. Within a name no spaces will appear. The next line contains one integer m, representing the length of the table to follow. The last m lines each contain the name ci of a source currency, a real number rij which represents the exchange rate from ci to cj and a name cj of the destination currency. Exchanges which do not appear in the table are impossible.
Test cases are separated from each other by a blank line. Input is terminated by a value of zero (0) for n.


Output
For each test case, print one line telling whether arbitrage is possible or not in the format "Case case: Yes" respectively "Case case: No".


Sample Input
3
USDollar
BritishPound
FrenchFranc
3
USDollar 0.5 BritishPound
BritishPound 10.0 FrenchFranc
FrenchFranc 0.21 USDollar

3
USDollar
BritishPound
FrenchFranc
6
USDollar 0.5 BritishPound
USDollar 4.9 FrenchFranc
BritishPound 10.0 FrenchFranc
BritishPound 1.99 USDollar
FrenchFranc 0.09 BritishPound
FrenchFranc 0.19 USDollar

0


Sample Output
Case 1: Yes
Case 2: No

求汇率，之间的汇率通过转换，如果能超过1，输出yes，否则输出no。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 #include<cmath>
 6 #include<map>
 7 using namespace std;
 8 
 9 struct Edge
10 {
11     int u,v;
12     double w;
13 }edge[10001];
14 double dis[10001];
15 
16 int n,m;
17 void bellmanford(int u0)
18 {
19     int i,j;
20     for(i=1;i<=n;i++)
21     {
22         dis[i]=0;
23     }
24     dis[u0]=1;
25     for(i=1;i<=n;i++)
26     {
27         for(j=1;j<=m;j++)
28         {
29             if(dis[edge[j].u]*edge[j].w>dis[edge[j].v])
30             {
31                 dis[edge[j].v]=dis[edge[j].u]*edge[j].w;
32             }
33         }
34     }
35 }
36 
37 int main()
38 {
39     int cas=1;
40     while(scanf("%d",&n)!=EOF&&n)
41     {
42         int i;
43         map<string,int>M;
44         M.clear();
45         char name[1001];
46         int flag=1;
47         for(i=1;i<=n;i++)
48         {
49             scanf("%s",name);
50             if(!M[name])
51             {
52                 M[name]=flag++;
53             }
54         }
55         char name1[1001];
56         char name2[1001];
57         double cc;
58         scanf("%d",&m);
59         for(i=1;i<=m;i++)
60         {
61             scanf("%s%lf%s",name1,&cc,name2);
62             edge[i].u=M[name1];
63             edge[i].v=M[name2];
64             edge[i].w=cc;
65         }
66         bool sign=0;
67         for(i=1;i<=n;i++)
68         {
69             bellmanford(i);
70             if(dis[i]>1.0)
71             {
72                 sign=1;break;
73             }
74         }
75         if(sign==1)
76         {
77             printf("Case %d: Yes\n",cas++);
78         }
79         else
80             printf("Case %d: No\n",cas++);
81     }
82     return 0;
83 }
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posted on 2013-07-24 23:24  张狂不年轻°  阅读(150)  评论(0)  编辑  收藏  举报