题意:给几个国家,然后给这些国家之间的汇率.判断能否通过这些汇率差进行套利交易. Floyd的算法可以求出任意两点间的最短路径,最后比较本国与本国的汇率差,如果大于1,则可以.否则不可以. 有向图 一个点到另一点的花费为权值相乘 求乘积的最大值 从点i出发 再回到点i的花费如果大于1 就可以 Sample Input3USDollarBritishPoundFrenchFranc3USDollar 0.5 BritishPoundBritishPound 10.0 FrenchFrancFren…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=1217 Arbitrage Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4430    Accepted Submission(s): 2013 Problem Description Arbitrage is the use of discr…

Arbitrage http://acm.hdu.edu.cn/showproblem.php?pid=1217 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…

POJ 2240 Arbitrage / ZOJ 1092 Arbitrage / HDU 1217 Arbitrage / SPOJ Arbitrage(图论,环) 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 exa…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1217 题目大意:问你是否可以通过转换货币从中获利 如下面这组样例: USDollar 0.5 BritishPound BritishPound 10.0 FrenchFranc FrenchFranc 0.21 USDollar 可以通过US->Br->French->US这样转换,把1美元变成1*0.5*10*0.21=1.05美元赚取%5的利润. 解题思路:其实就相当于bellman-…

给出一些国家之间的汇率,看看能否从中发现某些肮脏的......朋友交易. 这是Floyd的应用,dp思想,每次都选取最大值,最后看看自己跟自己的.....交易是否大于一.... #include<iostream> #include<cstring> #include<queue> #include<cstdio> #include<map> using namespace std; #define exp 0.00000001 map<s…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=1217 /************************************************************************/ /* hdu Arbitrage floyd求最短路径 题目大意:floyd算法,求得两点之间最短距离,(u,v) = min( (u,v),(u,w)+(w,v) ); 此题,是求其能够赚钱,即最大生成树,且过程是相乘的,公式:( u, v) =…

http://acm.hdu.edu.cn/showproblem.php?pid=1596 find the safest road Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6911    Accepted Submission(s): 2450 Problem Description XX星球有很多城市,每个城市之间有一条或…

http://acm.hdu.edu.cn/showproblem.php?pid=1869 六度分离 Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4355    Accepted Submission(s): 1768 Problem Description 1967年,美国著名的社会学家斯坦利·米尔格兰姆提出了一个名为“小世界现象…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3665 题意分析:以0为起点,求到Sea的最短路径. 所以可以N为超级汇点,使用floyd求0到N的最短路径. /*Seaside Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1151 Accepted Submission(s): 839 P…