poj 2240 Arbitrage
2016-05-24 09:06
281 查看
Arbitrage
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 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
Sample Output
Source
Ulm Local 1996
这是一道货币兑换,求是否存在正环的问题;
用BellmanFord算法可以解决;
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 19794 | Accepted: 8370 |
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 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
Source
Ulm Local 1996
这是一道货币兑换,求是否存在正环的问题;
用BellmanFord算法可以解决;
#include <iostream> #include <string> #include <cstring> #include <algorithm> #include <map> #include <cstdio> using namespace std; string currency[50]; map<string,int> mp; struct Edge { int from; int to; double weight; }edge[10000]; double dis[50]; int n,m; bool BellmanFord() { bool flag; for(int j=0;j<n;++j) { flag=false; for(int k=0;k<m;++k) { if(dis[edge[k].to]<dis[edge[k].from]*edge[k].weight) { dis[edge[k].to]=dis[edge[k].from]*edge[k].weight; flag=1; } } if(!flag) break; } for(int j=0;j<m;++j) { if(dis[edge[j].to]<dis[edge[j].from]*edge[j].weight) return true; } return false; } int main() { double weight; string a,b; int icase=0; while(cin>>n&&n) { for(int i=0;i<n;++i) { cin>>currency[i]; mp[currency[i]]=i; } cin>>m; for(int i=0;i<m;++i) { cin>>a>>weight>>b; edge[i].from=mp[a]; edge[i].to=mp[b]; edge[i].weight=weight; } memset(dis,0,sizeof(dis)); dis[0]=1; bool flag=BellmanFord(); printf("Case %d: ",++icase); if(flag) cout<<"Yes"<<endl; else cout<<"No"<<endl; } return 0; }
相关文章推荐
- 网页抓包实例---校园助手app
- Linux 免密码登录
- 计算机是如何启动的
- Java入门教程-1.1Java概述
- Leetcode 104. Maximum Depth of Binary Tree
- Android As报错:Warning:Gradle version 2.10 is required. Current version is 2.8. If using th....
- NYOJ 外星人的供给战--710
- spring的两大核心技术之一:控制反转
- 打包报错
- 通过Azure Service Fabric实现微服务
- 【iCore3 双核心板_ uC/OS-III】例程八:互斥信号量
- UIPickerView的使用
- Servlet运行机制与生命周期
- Ubuntu 16.04 LTS的这十项新功能,每个Ubuntu用户必须要知道!
- 强悍的 Linux —— 强悍的 vim (二)
- Comet4J 相关
- 关于登录模块的心得体会
- JAVA队列之优先队列
- 按两次返回键退出程序
- 动态规划详解