套汇问题,从源点做SPFA,如果有一个点入队次数大于v次(v表示点的个数)则图中存在负权回路,能够套汇,如果不存在负权回路,则判断下源点到自身的最长路是否大于自身,使用SPFA时松弛操作需要做调整 #include<iostream> #include<cstdio> #include<string.h> #include <stdlib.h> #include <math.h> using namespace std; const int ma…

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 same pair o…

Currency Exchange Time Limit:1000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 1860 Description Several currency exchange points are working in our city. Let us suppose that each point specializes in two par…

链接: http://poj.org/problem?id=1860 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22010#problem/A Currency Exchange Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 16244   Accepted: 5656 Description Several currency exchange point…

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…

传送门:http://poj.org/problem?id=1860 题意:给出每两种货币之间交换的手续费和汇率,求出从当前货币s开始交换回到s,能否使本金增多. 思路:bellman-Ford模板题.直接跑一遍,判断是否存在正环就好了.(复杂度n*m) 代码: #include<iostream> using namespace std; int n; //货币种数 int m; //兑换点数量 int s; //持有第s种货币 double v; //持有的s货币的本金 double di…

(￣▽￣)" #include<iostream> #include<cstdio> #include<cmath> #include<algorithm> #include<cstring> #include<vector> #include<queue> using namespace std; ; ; int N,M,S; double V,Vcur[MAXN],R[MAXN][MAXN],C[MAXN][…

一.题意 有多个货币交易点,每个只能互换两种货币,兑换的汇率不同,并收取相应的手续费.有N种货币,假定你拥有第S中,数量为V,有M个兑换点.问你能不能通过兑换操作使你最后拥有的S币比起始的时候多. 二.题解 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(在这一回路上,顶点的权值能不断增加,即能一直进行松弛),且最后得到的s金额是增加的.说到松弛我们立即会想到Bellman-Ford算法,它的特点是可以处理负权边,并能判断是否存在负环(负权回路).本题题恰恰与bellman-For…

POJ 1860 Currency Exchange / ZOJ 1544 Currency Exchange (最短路径相关,spfa求环) 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 onl…