hdu 1217(Floyed)
Arbitrage
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6522 Accepted Submission(s): 3019
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 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.
each test case, print one line telling whether arbitrage is possible or
not in the format "Case case: Yes" respectively "Case case: No".
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
Case 2: No
import java.util.HashMap;
import java.util.Scanner; public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = ;
while(sc.hasNext()){
int n = sc.nextInt();
if(n==) break;
double [][] map = new double [n][n];
HashMap<String,Integer> hm = new HashMap<String,Integer>();
for(int i=;i<n;i++){
String str = sc.next();
hm.put(str,i);
}
int m = sc.nextInt();
while(m-->){
String str1 = sc.next();
double v = sc.nextDouble();
String str2 = sc.next();
int a = hm.get(str1);
int b = hm.get(str2);
map[a][b]= v;
}
for(int k=;k<n;k++){
for(int i=;i<n;i++){
for(int j=;j<n;j++){
if(map[i][j]<map[i][k]*map[k][j]) map[i][j] = map[i][k]*map[k][j];
}
}
}
boolean flag = false;
for(int i=;i<n;i++) if(map[i][i]>){flag=true;break;}
System.out.print("Case "+(t++)+": ");
if(flag) System.out.println("Yes");
else System.out.println("No");
}
}
}
hdu 1217(Floyed)的更多相关文章
- POJ 2240 Arbitrage / ZOJ 1092 Arbitrage / HDU 1217 Arbitrage / SPOJ Arbitrage(图论,环)
POJ 2240 Arbitrage / ZOJ 1092 Arbitrage / HDU 1217 Arbitrage / SPOJ Arbitrage(图论,环) Description Arbi ...
- [ACM] hdu 1217 Arbitrage (bellman_ford最短路,推断是否有正权回路或Floyed)
Arbitrage Problem Description Arbitrage is the use of discrepancies in currency exchange rates to tr ...
- hdu 1217 (Floyd变形)
链接:http://acm.hdu.edu.cn/showproblem.php?pid=1217 Arbitrage Time Limit: 2000/1000 MS (Java/Others) ...
- HDU 1217 Arbitrage (Floyd)
Arbitrage http://acm.hdu.edu.cn/showproblem.php?pid=1217 Problem Description Arbitrage is the use of ...
- hdu 1217 Arbitrage (最小生成树)
题目:http://acm.hdu.edu.cn/showproblem.php?pid=1217 /************************************************* ...
- HDU 1217 Arbitrage(Bellman-Ford判断负环+Floyd)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1217 题目大意:问你是否可以通过转换货币从中获利 如下面这组样例: USDollar 0.5 Brit ...
- hdu 1217 Arbitrage (spfa算法)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1217 题目大意:通过货币的转换,来判断是否获利,如果获利则输出Yes,否则输出No. 这里介绍一个ST ...
- HDOJ 1217 Floyed Template
解题思路:1.map简单应用2.Floyd算法的变形,之后判断dis[i][i],如果大于1,则存在利润! #include <iostream> #include <stdio.h ...
- hdu 1217 汇率 Floyd
题意:给几个国家,然后给这些国家之间的汇率.判断能否通过这些汇率差进行套利交易. Floyd的算法可以求出任意两点间的最短路径,最后比较本国与本国的汇率差,如果大于1,则可以.否则不可以. 有向图 一 ...
随机推荐
- tools.jar seem to ....
android stadio 运行不起来,可以在java_home原来的路径下加一个\, 然后就可以运行起来了.
- css一些事儿
1. margin和padding 如果边界画一条线,则margin的属于边界外,padding属于边界内 当我们给元素背景色时,margin区域不会被着色,而padding区域会被着色. 当上下两个 ...
- 网易OpenStack部署运维实战
OpenStack自2010年项目成立以来,已经有超过200个公司加入了 OpenStack 项目,目前参与 OpenStack 项目的开发人员有 17,000+,而且这些数字还在增加,作为一个开源的 ...
- VSX-1 概述
博客搁置了一段时间,一直想写,无从下手,正好最近在做VS2010扩展方面的项目,所以写VSX系列文章以记之. 背景 现有工作是做金融行业,主要项目是一套银行综合前端系统,也就是银行平时用的最多的一个系 ...
- Django笔记 —— 模板
最近在学习Django,打算玩玩网页后台方面的东西,因为一直很好奇但却没怎么接触过.Django对我来说是一个全新的内容,思路想来也是全新的,或许并不能写得很明白,所以大家就凑合着看吧- 本篇笔记(其 ...
- 自动化测试学习之路--HTML常见元素、属性的简单学习
如何创建html文件: 使用工具:VSCode 1.双击文件名显示区,可快速新建文件. 2.保存文件,文件名.html 3.输入!(必须是英文的!),按 Tab键,可自动生成html格式的文件,如下: ...
- day-python入门3
本节内容 鸡汤.电影 IDE介绍 知识回顾 数据类型 For循环 while循环 列表及常用操作 IDE介绍 IDE即集成开发环境 常见IDE Visualstudio : w ...
- win10 64位 C# 连接oracle 32位, 遇到的问题及解决
首次 本机电脑是win10系统 64位的:安装的oracle数据库也是64位的: 服务器端的oracle 是32位的: 第一次安装的pl/sql 也是64位的, 配置完 F:\app\ln_qi\p ...
- ASP.NET Core API ---状态码
摘录自:https://www.cnblogs.com/cgzl/p/9047626.html 状态码是非常重要的,因为只有状态码会告诉API的消费者: 请求是否如预期的成功,或者失败 如果出现了错误 ...
- ASP.net MVC入门及Razor语法
一.MVC入门: 1.MVC简介 约定大于配置 2.MVC访问流程 csthml模板(razor模板)就是简化HTML的拼接的模板,最终还是生成html给浏览器显示,不能直接访问cshtml文件. 二 ...