poj 2240 Arbitrage
来源:互联网 发布:linux ssh ip port 编辑:程序博客网 时间:2024/05/16 11:05
Arbitrage
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 19794 Accepted: 8370
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.
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.
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
3USDollarBritishPoundFrenchFranc3USDollar 0.5 BritishPoundBritishPound 10.0 FrenchFrancFrenchFranc 0.21 USDollar3USDollarBritishPoundFrenchFranc6USDollar 0.5 BritishPoundUSDollar 4.9 FrenchFrancBritishPound 10.0 FrenchFrancBritishPound 1.99 USDollarFrenchFranc 0.09 BritishPoundFrenchFranc 0.19 USDollar0
Sample Output
Case 1: YesCase 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;}
0 0
- POJ 2240 Arbitrage
- poj 2240 Arbitrage
- POJ 2240 Arbitrage (Floyd)
- poj 2240Arbitrage(Floyd)
- POJ 2240 Arbitrage Floyd
- Poj 2240 Arbitrage
- POJ 2240 Arbitrage
- poj 2240 Arbitrage
- POJ 2240 Arbitrage (spfa)
- POJ 2240 Arbitrage
- POJ 2240 Arbitrage
- Poj 2240 Arbitrage
- POJ 2240 Arbitrage
- POJ 2240 Arbitrage
- POJ 2240 Arbitrage
- POJ 2240 Arbitrage
- POJ-2240-Arbitrage
- poj 2240 Arbitrage
- Ubuntu 16.04 LTS的这十项新功能,每个Ubuntu用户必须要知道!
- UIPickerView的使用
- Servlet运行机制与生命周期
- 控制台里打印view的层级 在控制台里打印controller的层级 及其他技巧
- 打包报错
- poj 2240 Arbitrage
- NYOJ 外星人的供给战--710
- Android As报错:Warning:Gradle version 2.10 is required. Current version is 2.8. If using th....
- Leetcode 104. Maximum Depth of Binary Tree
- Java入门教程-1.1Java概述
- 计算机是如何启动的
- Linux 免密码登录
- 网页抓包实例---校园助手app
- selectNode、selectNodeContents 区别