POJ1860:Currency Exchange(BF)】的更多相关文章

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…
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…
链接:http://poj.org/problem?id=1860 Currency Exchange 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 currenc…
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…
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…
题目链接:http://poj.org/problem?id=1860 Currency Exchange Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 32128   Accepted: 12228 Description Several currency exchange points are working in our city. Let us suppose that each point specialize…
题目链接: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 b…
题目链接:http://poj.org/problem?id=1860 题目大意:给你一些兑换方式,问你能否通过换钱来赚钱? 使用ford算法,当出现赚钱的时候就返回YES,如果不能赚钱,则返回NO 应该是可以停下来的,但是我不会分析复杂度,谁来教教我? #include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <map> #inclu…
题目链接. 分析: 以前没做出来,今天看了一遍题竟然直接A了.出乎意料. 大意是这样,给定不同的金币的编号,以及他们之间的汇率.手续费,求有没有可能通过不断转换而盈利. 直接用Bellman-ford检测负环的方法检测. #include <iostream> #include <cstdio> #include <cstring> using namespace std; ; <<); struct Node { int u, v; double r, c…
题意:有m个货币交换点,每个点只能有两种货币的互相交换,且要给佣金,给定一开始的货币类型和货币数量,问若干次交换后能否让钱增加. 思路:spfa求最长路,判断是否存在正环,如果存在则钱可以在环中一直增加,最后的钱肯定也是增加的. #include <iostream> #include <cstring> #include <queue> #include <cstdio> using namespace std; + ; struct edge{ int…