POJ 2240 Arbitrage
来源:互联网 发布:iapp源码手册2.0 编辑:程序博客网 时间:2024/06/16 03:56
bellmanford最长路判正环
Arbitrage
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 13947 Accepted: 5876
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
#include <iostream>#include <cstdio>#include <cstring>#include <map>#include <cmath>#include <algorithm>using namespace std;const double eps=1e-8;double g[120][120];int n,m;double dist[120];map<string,int> mp;bool Bellman_Ford(){ memset(dist,0,sizeof(dist)); dist[0]=1; for(int i=0;i<n-1;i++) { bool flag=false; for(int j=0;j<n;j++) { for(int k=0;k<n;k++) { if(fabs(g[j][k])>eps) { if(dist[k]<dist[j]*g[j][k]) { flag=true; dist[k]=dist[j]*g[j][k]; } } } } if(!flag) return false; } bool flag=false; for(int i=0;i<n;i++) { for(int j=0;j<n;j++) { if(dist[i]<dist[j]*g[j][i]) { flag=true; } if(flag) break; } if(flag) break; } return flag;}int main(){ int cas=1;while(scanf("%d",&n)!=EOF&&n){ mp.clear(); for(int i=0;i<n;i++) { string name; cin>>name; mp[name]=i; } memset(g,0,sizeof(g)); scanf("%d",&m); while(m--) { string name1,name2; double rate; cin>>name1>>rate>>name2; g[mp[name1]][mp[name2]]=rate; } printf("Case %d: ",cas++); if(Bellman_Ford()) puts("Yes"); else puts("No");} return 0;}
1 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
- 什么是内存泄漏
- 排序系列之快速排序
- 虚拟机下装oracle10G ASM
- UVa:11995 I Can Guess the Data Structure!
- 集合类接口和类层次关系图
- POJ 2240 Arbitrage
- Yii Framework 开发教程(34) Zii组件-AutoComplete示例
- 一个简单的EBNF范式的实现
- Yii Framework 开发教程(35) Zii组件-Button示例
- android 点击效果动画增强
- redhat 常用命令备忘
- 各类女性,各类人生
- SVN服务器搭建和使用(一)
- [Leetcode] Maximum Depth of Binary Tree (Java)