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 pound buys 10.0 Fre…

Arbitrage 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 pound buys 10.0 Frenc…

d[i]代表从起点出发可以获得最多的钱数,松弛是d[v]=r*d[u],求最长路,看有没有正环 然后这题输入有毒,千万别用cin 因为是大输入,组数比较多,然后找字符串用strcmp就好,千万不要用map 这题刚开始我T了(用的map),还以为组数很多卡spfa呢,然后我上网看了看都是floyd的,然后我用floyd写了一发,891ms过了 然后我感觉spfa的复杂度也不是很大,就是看有没有正环,所以我觉得可能是map+cin的锅,然后改了一发,用的spfa,47ms过 真是,算了,实质是本蒟蒻…

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…

题目传送门 /* 最短路:Floyd模板题 只要把+改为*就ok了,热闹后判断d[i][i]是否大于1 文件输入的ONLINE_JUDGE少写了个_,WA了N遍:) */ #include <cstdio> #include <iostream> #include <cstring> #include <algorithm> #include <string> #include <map> #include <cmath>…

Arbitrage Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 21300   Accepted: 9079 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 currenc…

链接:poj 2240 题意:首先给出N中货币,然后给出了这N种货币之间的兑换的兑换率. 如 USDollar 0.5 BritishPound 表示 :1 USDollar兑换成0.5 BritishPound. 问在这N种货币中是否存在货币经过若干次兑换后,兑换成原来的货币能够使货币量添加. 思路:本题事实上是Floyd的变形.将变换率作为构成图的路径的权值.只是构成的图是一个有向图. 最后将松弛操作变换为:if(dis[i][j]<dis[i][k]*dis[k][j]). #includ…

(￣▽￣)" #include<iostream> #include<cstdio> #include<cmath> #include<algorithm> #include<cstring> #include<string> #include<vector> #include<queue> using namespace std; ; int n,m; double Vcur[MAXN],R[MAX…

链接: http://poj.org/problem?id=2240 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22010#problem/F Arbitrage Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13067   Accepted: 5493 Description Arbitrage is the use of discrepancies i…

点击打开链接 Arbitrage Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13434   Accepted: 5657 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…

http://poj.org/problem?id=2240 题意 : 好吧,又是一个换钱的题:套利是利用货币汇率的差异进行的货币转换,例如用1美元购买0.5英镑,1英镑可以购买10法郎,一法郎可以购买0.21美元,所以0.5*10*0.21 = 1.05,从中获利百分之五,所以需要编写一个程序,在进行完转换之后能不能获利,如果能就输出Yes,反之No: 样例解释 : 3 USDollar BritishPound FrenchFranc 6 USDollar 0.5 BritishPound…

Arbitrage 题目链接: http://acm.hust.edu.cn/vjudge/contest/122685#problem/I 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…

题目:http://poj.org/problem?id=2240 题意:给定n个货币名称,给m个货币之间的汇率,求会不会增加 和1860差不多,求有没有正环 刚开始没对,不知道为什么用 double往结构体里传值的时候 会去掉小数点后的 数 #include <iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<stack> #include<qu…

http://poj.org/problem?id=2240 题意:货币兑换,判断最否是否能获利. 思路:又是货币兑换题,Belloman-ford和floyd算法都可以的. #include<iostream> #include<algorithm> #include<string> #include<cstring> #include<map> using namespace std; + ; int n, m; string s1,s2;…

题目链接:http://poj.org/problem?id=2240 题目就是要通过还钱涨自己的本钱最后还能换回到自己原来的钱种. 就是判一下有没有负环那么就直接用bellman_ford来判断有没有负环 #include <iostream> #include <cstring> #include <string> using namespace std; int n , m , s , a , b , counts; double rx , ry , cx , c…

Time Limit: 1000 MS Memory Limit: 65536 KB 64-bit integer IO format: %I64d , %I64u   Java class name: Main [Submit] [Status] [Discuss] Description Arbitrage is the use of discrepancies in currency exchange rates to transform one unit of a currency in…

题意  给你n种币种之间的汇率关系  推断是否能形成套汇现象  即某币种多次换为其他币种再换回来结果比原来多 基础的最短路  仅仅是加号换为了乘号 #include<cstdio> #include<cstring> #include<string> #include<map> using namespace std; map<string, int> na; const int N = 31; double d[N], rate[N][N],…

Arbitrage Time Limit: 1000MS   Memory Limit: 65536K Total Submissions:27167   Accepted: 11440 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 currenc…

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 pound buys 10.0 Fre…

Time Limit: 1000MS Memory Limit: 65536K 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 Brit…

Arbitrage Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13800   Accepted: 5815 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 currenc…

题意: 给出一些货币和货币之间的兑换比率,问是否可以使某种货币经过一些列兑换之后,货币值增加. 举例说就是1美元经过一些兑换之后,超过1美元.可以输出Yes,否则输出No. 分析: 首先我们要把货币之间的关系转化成一张图.转化时,用STL里面的map很方便. 为每种货币分配一个序列号,一个序列号代表了一个图中间的NODE,而node之间的edge用汇率表示. 一开始用Dijkstra算法做,死活AC不了,网友给的理由是: 由于Dijkstra算法不能处理带有负权值的最短路,但此题中,两种货币之间…

和POJ1860差不多,就是用bellmanford判读是否存在正环,注意的是同种货币之间也可以交换,就是说:A货币换A货币汇率是2的情况也是存在的. #include<stdio.h> #include<string.h> #include<cstring> #include<string> #include<math.h> #include<queue> #include<algorithm> #include<…

https://vjudge.net/problem/POJ-2240 题意 已知n种货币,以及m种货币汇率及方式,问能否通过货币转换,使得财富增加. 分析 Bellman-Ford判断正环,注意初始化时置为0. #include<cstdio> #include<cstring> #include<string> #include<algorithm> #include<iostream> #include<vector> #inc…

题意:给出n种货币,m种兑换比率(一种货币兑换为另一种货币的比率),判断测试用例中套汇是否可行.(套汇的意思就是在经过一系列的货币兑换之后,是否可以获利.例如:货币i→货币j→货币i,这样兑换后,是否可以获利,即比率是否>1).举个例子理解套汇:假设,1美元买10人民币(比率为10),10人民币买100美元(比率为10).这就是套汇,总比率=10*10=100,100>1,所以可以获利. 思路:Floyed-Warshall算法,枚举所有的货币之间的兑换比率,最后若存在一种回路且回路比率>…

三道题都是考察最短路算法的判环.其中1860和2240判断正环,3259判断负环. 难度都不大,可以使用Bellman-ford算法,或者SPFA算法.也有用弗洛伊德算法的,笔者还不会SF-_-…… 直接贴代码. 1860 Currency Exchange: #include <cstdio> #include <cstring> int N,M,S; double V; ; int first[maxn],vv[maxn*maxn],nxt[maxn*maxn]; double…

http://poj.org/problem?id=2240 用log化乘法为加法找正圈 c++ 110ms,g++tle #include <string> #include <map> #include <iostream> #include <cmath> #include <cstring> #include <queue> using namespace std; const int maxn = 50; bool vis[…

  Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 17374   Accepted: 7312 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 e…

题目链接:http://poj.org/problem?id=2240 题意:n种国家的货币,m个换算汇率.问你能不能赚钱. eg:1美元换0.5英镑,1英镑换10法郎,1法郎换0.21美元,这样1美元能换0.5*10*0.21=1.05美元,净赚0.05美元.   题解:floyd跑的时候改成乘法运算.然后就是在货币上做一个map的映射. 代码: #include <iostream> #include <string> #include <map> #include…

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…