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…

Currency Exchange POJ - 1860 题意: 有许多货币兑换点,每个兑换点仅支持两种货币的兑换,兑换有相应的汇率和手续费.你有s这个货币 V 个,问是否能通过合理地兑换货币,使得你手中的货币折合成s后是有增加的. 思路: 这道题在建立每种货币的兑换关系后,找到图中的正环即可,因为你沿着正环跑就可以增加价值.这里可以用类似Bellman_Ford判断负环的方法. #include <algorithm> #include <iterator> #include &…

转载来源:優YoU  http://user.qzone.qq.com/289065406/blog/1299337940 提示:关键在于反向利用Bellman-Ford算法 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,…

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…

//spfa 判断正环 #include<iostream> #include<queue> #include<cstring> using namespace std; const int N=1e4; const int INF=2e9; int h[N],to[N],ne[N],idx; double r[N],c[N]; int n, m,X; double V; void add(int u,int v,double r1,double c1) { to[id…

Farmer John has decided to reward his cows for their hard work by taking them on a tour of the big city! The cows must decide how best to spend their free time. Fortunately, they have a detailed city map showing the L (2 ≤ L ≤ 1000) major landmarks (…

XYZZY Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4421    Accepted Submission(s): 1252 Problem Description It has recently been discovered how to run open-source software on the Y-Crate gami…

XYZZY Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5304    Accepted Submission(s): 1510 Problem Description It has recently been discovered how to run open-source software on the Y-Crate gam…