//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…

题意:给出n种货币,m中交换关系,给出两种货币汇率和手续费,求能不能通过货币间的兑换使财富增加. 用Bellman_Ford 求出是否有正环,如果有的话就可以无限水松弛,财富可以无限增加. #include<string.h> #include<stdio.h> const int N=110; const int inf=0x3fffffff; int start,num,n; double dist[N],wf; struct edge { int st,ed; double…

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…

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…

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…

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…

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>…

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…

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: http://acm.hdu.edu.cn/showproblem.php?pid=1317 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=29015#problem/F XYZZY Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1701    Accepted S…

传送门:点击打开链接 题目大意:一个城市有n种货币,m个货币交换点,你有v的钱,每个交换点只能交换两种货币,(A换B或者B换A),每一次交换都有独特的汇率和手续费,问你存不存在一种换法使原来的钱更多. 思路:一开始以为一个地方只能用一次,感觉好像有点难,后来发现自己读错题了,其实只要判断给你的这幅图存不存在正环就可以了,用dis[]表示某种货币的数量,然后bellman判断正环就可以了.(题目里强调结尾一定要原来的货币,但其实这是废话,因为是以原来的货币为起点的,所以你换出去了一定换的回来),正…

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…

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…

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<…

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 &…

XYZZY Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 4154   Accepted: 1185 Description The prototypical computer adventure game, first designed by Will Crowther on the PDP-10 in the mid-1970s as an attempt at computer-refereed fantasy g…

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…

题目链接: https://vjudge.net/problem/POJ-2240 题目大意: 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. 思路: 由于这里问的是财富有没有增加,但是没有源点,所以可以枚举1-n为源点,分别用bellman-ford算法判断是否存在正环,如果有正环那就输出Yes,反之输出No. 这里还需要编号 其实可以用Floyd算法直接出结果判断有没有正环,这里用bellman-ford是卡着时间过的. #include<iostream> #i…

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…

心累,陕西邀请赛学校不支持,可能要自费了.. 思路:套用Bellman-Ford判断负环的思路,把大于改成小于即可判定是否存在从源点能到达的正环.如果存在正环,那么完全多跑几次正环就可以把钱增加到足够返回到S并且大于原来的金额. AC代码 #include <cstdio> #include <cmath> #include <cctype> #include <algorithm> #include <cstring> #include <…

#include<stdio.h> #include<string.h> #include<queue>//只需判断是否有正环路径就可以了 using namespace std; #define N  200 struct node { double r,c; }map[N][N]; double maxvalue[N],h; int n,cou[N]; int  bellmanford(int start) {   queue<int>q;   int…

//判断负环 dist初始化为正无穷 //正环 负无穷 #include<iostream> #include<cstring> #include<queue> #include<algorithm> using namespace std; , M = ; int dist[N]; int h[N], e[M], w[M], ne[M], idx; int n,m,z; void add(int a,int b,int c) { e[idx]=b; w[i…

题意:给出n个房间,初始在房间1有100的能量值,每次进入一个房间,能量值可能增加也可能减小,(是点权,不是边权),问能否到达终点的时候能量值还为正 这题自己写的时候wa--wa-- 后来看了题解,还是wa---wa--- 题解很详细http://blog.csdn.net/freezhanacmore/article/details/9937327 记录下自己犯的错误吧 首先是floyd函数初始化的时候,直接写在了函数里面,这样是不对的,因为输入值在前,调用函数在后,这样就相当于将之前输入的可…

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…

题目链接: https://vjudge.net/problem/POJ-1860 题目大意: 我们的城市有几个货币兑换点.让我们假设每一个点都只能兑换专门的两种货币.可以有几个点,专门从事相同货币兑换.每个点都有自己的汇率,外汇汇率的A到B是B的数量你1A.同时各交换点有一些佣金,你要为你的交换操作的总和.在来源货币中总是收取佣金. 例如,如果你想换100美元到俄罗斯卢布兑换点,那里的汇率是29.75,而佣金是0.39,你会得到(100 - 0.39)×29.75＝2963.3975卢布. 你…

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…

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 (…

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;…

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…