Arbitrage
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 16652   Accepted: 7004

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

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

 #include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
const int M = , inf = 0x3f3f3f3f;
struct Arbitrage
{
int u , v ;
double r ;
}e[M * M];
int n , m ;
char cur[M][] ;
char a[] , b[] ;
double d[M] ; void init (char a[] , char b[] , int no)
{
for (int i = ; i <= n ; i++) {
if (strcmp (cur[i] , a) == )
e[no].u = i ;
if (strcmp (cur[i] , b) == )
e[no].v = i ;
}
} int bellman_ford (int o)
{
for (int i = ; i <= n ; i++)
d[i] = ;
d[o] = 1.0 ;
double temp = 1.0 ;
bool flag ;
for (int i = ; i <= n ; i++) {
flag = ;
for (int j = ; j < m ; j++) {
if (d[e[j].v] < d[e[j].u] * e[j].r) {
d[e[j].v] = d[e[j].u] * e[j].r ;
flag = ;
}
if (d[o] > temp) {
return true ;
}
}
}
return false ;
} int main ()
{
//freopen ("a.txt" , "r" , stdin) ;
int ans = ;
while (~ scanf ("%d" , &n)) {
if (n == )
break ;
getchar () ;
for (int i = ; i <= n ; i++) {
gets (cur[i]) ;
}
scanf ("%d" , &m) ;
for (int i = ; i < m ; i++) {
cin >> a >> e[i].r >> b ;
init (a , b , i) ;
}
/* for (int i = 0 ; i < m ; i++) {
printf ("u = %d , v = %d , r = %.2f\n" , e[i].u , e[i].v , e[i].r) ;
}*/
int i ;
for (i = ; i <= n ; i++) {
if (bellman_ford (i)) {
printf ("Case %d: Yes\n" , ans++) ;
// printf ("Arbitrage num : %d\n" , i) ;
break ;
}
/*printf ("%d team: \n" , i) ;
for (int j = 1 ; j <= n ; j++)
printf ("%.2f " ,d[j]) ;
puts ("") ; */
}
if (i == n + )
printf ("Case %d: No\n" , ans++) ;
}
return ;
}

Arbitrage(bellman_ford)的更多相关文章

  1. [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 ...

  2. POJ 2240 Arbitrage Bellman_ford 判读是否存在正环

    和POJ1860差不多,就是用bellmanford判读是否存在正环,注意的是同种货币之间也可以交换,就是说:A货币换A货币汇率是2的情况也是存在的. #include<stdio.h> ...

  3. POJ 2240 Arbitrage【Bellman_ford坑】

    链接: http://poj.org/problem?id=2240 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22010#probl ...

  4. POJ 2240 - Arbitrage(bellman_ford & floyd)

    题意: 给出一些货币和货币之间的兑换比率,问是否可以使某种货币经过一些列兑换之后,货币值增加. 举例说就是1美元经过一些兑换之后,超过1美元.可以输出Yes,否则输出No. 分析: 首先我们要把货币之 ...

  5. poj 2240 Arbitrage (最短路 bellman_ford)

    题目:http://poj.org/problem?id=2240 题意:给定n个货币名称,给m个货币之间的汇率,求会不会增加 和1860差不多,求有没有正环 刚开始没对,不知道为什么用 double ...

  6. poj 2240 Arbitrage(Bellman_ford变形)

    题目链接:http://poj.org/problem?id=2240 题目就是要通过还钱涨自己的本钱最后还能换回到自己原来的钱种. 就是判一下有没有负环那么就直接用bellman_ford来判断有没 ...

  7. POJ 1860 Currency Exchange + 2240 Arbitrage + 3259 Wormholes 解题报告

    三道题都是考察最短路算法的判环.其中1860和2240判断正环,3259判断负环. 难度都不大,可以使用Bellman-ford算法,或者SPFA算法.也有用弗洛伊德算法的,笔者还不会SF-_-…… ...

  8. Nyoj Arbitrage(Floyd or spfa or Bellman-Ford)

    描述Arbitrage is the use of discrepancies in currency exchange rates to transform one unit of a curren ...

  9. POJ 2240 - Arbitrage - [bellman-ford求最短路]

    Time Limit: 1000MS Memory Limit: 65536K Description Arbitrage is the use of discrepancies in currenc ...

随机推荐

  1. SpringMVC类型转换、数据绑定详解[附带源码分析]

    目录 前言 属性编辑器介绍 重要接口和类介绍 部分类和接口测试 源码分析 编写自定义的属性编辑器 总结 参考资料 前言 SpringMVC是目前主流的Web MVC框架之一. 如果有同学对它不熟悉,那 ...

  2. C# 有关命名法

    常用的几种:匈牙利命名法,骆驼命名法,帕斯卡(pascal)命名法就说了. 总结下VS所用的标准命名: 1 ADO.NET 命名规范 数据类型 数据类型简写 标准命名举例 Connection con ...

  3. display:inline-block的坑

    一直用display:inline-block做某种导航栏还很爽,突然有一个柱状图的需求便也这么做了,于是成功被坑. 简简单单个需求,大致这样 只用几个li加上display:inline-block ...

  4. [C#基础]ref和out的区别

    在C#中通过使用方法来获取返回值时,通常只能得到一个返回值.因此,当一个方法需要返回多个值的时候,就需要用到ref和out,那么这两个方法区别在哪儿呢? MSDN:       ref 关键字使参数按 ...

  5. 第四十二课:基于CSS的动画引擎

    由于低版本浏览器不支持css3 animation,因此我们需要根据浏览器来选择不同的动画引擎.如果浏览器支持css3 animation,那么就使用此动画引擎,如果不支持,就使用javascript ...

  6. JavaScript事件冒泡简介及应用

    一.什么是事件冒泡 在一个对象上触发某类事件(比如单击onclick事件),如果此对象定义了此事件的处理程序,那么此事件就会调用这个处理程序,如果没有定义此事 件处理程序或者事件返回true,那么这个 ...

  7. Jetty+json-lib库抛异常的问题解决过程(java.lang.NoClassDefFoundError: net/sf/json/JSONObject)

      一.之前抛异常是将json库改成了fastjson解决的,参见: http://www.cnblogs.com/gossip/p/5369670.html     异常信息:     二.解决步骤 ...

  8. 关于obj和基本类通过函数参数传进去执行是否改变原来的值

    var obj = { p1 : 1, p2 : 2 }; (function(_/* 这个东东是地址的应用哦 */){ _.p1 = 3, _.p2 = 4 })(obj) var i = 2; ( ...

  9. Java-TreeSet

    如下: package 集合类.Set类; /** * Set不允许重复数据 */ /** * TreeSet 是用来进行集合排序的,请注意他和LinkedHashSet的区别. TreeSet是按照 ...

  10. Session的异常

    既然这一天就这么废了,那就多说一些吧!其实session也是有潜在的问题的.Session销毁的三种情况: (1)超时:超过30分钟 (2)服务器非正常关闭,如果自己手动stop service而不是 ...