<题目链接> <转载于 >>> > 题目大意: 有多种汇币,汇币之间可以交换,这需要手续费,当你用100A币交换B币时,A到B的汇率是29.75,手续费是0.39,那么你可以得到(100 - 0.39) * 29.75 = 2963.3975 B币.问s币的金额经过交换最终得到的s币金额数能否增加. 货币的交换是可以重复多次的,所以我们需要找出是否存在正权回路,且最后得到的s金额是增加的 怎么找正权回路呢?(正权回路:在这一回路上,顶点的权值能不断增加即能一直进行…

Currency Exchange Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 19881   Accepted: 7114 Description Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and pe…

Description Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operations only with these currencies. There can be several points specializing in the…

链接:poj 1860 题意:给定n中货币.以及它们之间的税率.A货币转化为B货币的公式为 B=(V-Cab)*Rab,当中V为A的货币量, 求货币S通过若干此转换,再转换为原本的货币时是否会添加 分析:这个题就是推断是否存在正权回路.能够用bellman-ford算法,只是松弛条件相反 也能够用SPFA算法,推断经过转换后,转换为原本货币的值是否比原值大... bellman-ford    0MS #include<stdio.h> #include<string.h> str…

题意:n种钱,m种汇率转换,若ab汇率p,手续费q,则b=(a-q)*p,你有第s种钱v数量,问你能不能通过转化让你的s种钱变多? 思路:因为过程中可能有负权值,用spfa.求是否有正权回路,dis[s]是否增加.把dis初始化为0,然后转化,如果能增大就更新.每次都判断一下dis[s]. 参考:最快最好用的——spfa算法 代码: #include<cstdio> #include<set> #include<vector> #include<cmath>…

Currency Exchange DescriptionSeveral currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operations only with these currencies. There can be several points sp…

Arbitrage Problem 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 p…

http://poj.org/problem?id=1860 Description Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operations only with these currencies. There can be sev…

POJ1860 题目大意:你在某一点有一些钱,给定你两点之间钱得兑换规则,问你有没有办法使你手里的钱增多.就是想看看转一圈我的钱能不能增多,出现这一点得条件就是有兑换钱得正权回路,所以选择用bellman_ford得算法 准备工作,bellman_ford得更新是根据已知边得数据来的,所以需要存住边的数据,还要有dis数组作为更新判断,dis也就相当于中心节点得钱兑换到i节点(中间可能会经过别的)变成了多少钱 #include <iostream> #include <string.h&…

题意:有最多一百个房间,房间之间连通,到达另一个房间会消耗能量值或者增加能量值,求是否能从一号房间到达n号房间. 看数据,有定5个房间,下面有5行,第 iii 行代表 iii 号 房间的信息,第一个数字表示从此房间到达连接的房间得到的能量,第二个数字表示连接的有几个房间,后面输出房间后. 思路: 正向去模拟,求出到达n点后尽可能的让dis[n]的值更大 , dis[1]初始化为100,其他初始化为零,因为在松弛的时候必须保证能量值大于零. 这里面说一下SPFA正权回路的判断,当一个点进入队列第n…