POJ 1860 Currency Exchange (Bellman ford)
2015-07-25 23:41
357 查看
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 currencies. There can be several points specializing in the same pair of currencies. Each point has its own exchange rates, exchange rate of A to B is the quantity of B you get for 1A. Also each exchange point has some commission, the sum you have to pay for your exchange operation. Commission is always collected in source currency.
For example, if you want to exchange 100 US Dollars into Russian Rubles at the exchange point, where the exchange rate is 29.75, and the commission is 0.39 you will get (100 - 0.39) * 29.75 = 2963.3975RUR.
You surely know that there are N different currencies you can deal with in our city. Let us assign unique integer number from 1 to N to each currency. Then each exchange point can be described with 6 numbers: integer A and B - numbers of currencies it exchanges, and real RAB, CAB, RBA and CBA - exchange rates and commissions when exchanging A to B and B to A respectively.
Nick has some money in currency S and wonders if he can somehow, after some exchange operations, increase his capital. Of course, he wants to have his money in currency S in the end. Help him to answer this difficult question. Nick must always have non-negative sum of money while making his operations.
Input
The first line of the input contains four numbers: N - the number of currencies, M - the number of exchange points, S - the number of currency Nick has and V - the quantity of currency units he has. The following M lines contain 6 numbers each - the description of the corresponding exchange point - in specified above order. Numbers are separated by one or more spaces. 1<=S<=N<=100, 1<=M<=100, V is real number, 0<=V<=103.
For each point exchange rates and commissions are real, given with at most two digits after the decimal point, 10-2<=rate<=102, 0<=commission<=102.
Let us call some sequence of the exchange operations simple if no exchange point is used more than once in this sequence. You may assume that ratio of the numeric values of the sums at the end and at the beginning of any simple sequence of the exchange operations will be less than 104.
Output
If Nick can increase his wealth, output YES, in other case output NO to the output file.
Sample Input
Sample Output
Source
Northeastern Europe 2001, Northern Subregion
借鉴kuangbin大大/article/4680056.html
题意:给定M种货币,有N个兑换点,开始手中有的货币种类是S并且数量为V,问能否从起点出发经过货币的兑换后再回到起点时,货币S的数量会增加
用100A兑换B时,汇率为29.75,手续费为0.39,那么最后得到的B为(100 - 0.39) * 29.75 = 2963.39
分析:是否存在正权回路,使得回路上的顶点的权值能够不断进行松弛
采用Bellman ford算法
原本的Bellman ford算法可以判断是否存在负圈,那么这题就是反向运用Bellman ford
如果存在S到其他点的距离能够不断变大时,存在正权回路
初始化d[S]=V 而源点到其他点的距离初始化为无穷小(0)
View Code
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 22405 | Accepted: 8095 |
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 of currencies. Each point has its own exchange rates, exchange rate of A to B is the quantity of B you get for 1A. Also each exchange point has some commission, the sum you have to pay for your exchange operation. Commission is always collected in source currency.
For example, if you want to exchange 100 US Dollars into Russian Rubles at the exchange point, where the exchange rate is 29.75, and the commission is 0.39 you will get (100 - 0.39) * 29.75 = 2963.3975RUR.
You surely know that there are N different currencies you can deal with in our city. Let us assign unique integer number from 1 to N to each currency. Then each exchange point can be described with 6 numbers: integer A and B - numbers of currencies it exchanges, and real RAB, CAB, RBA and CBA - exchange rates and commissions when exchanging A to B and B to A respectively.
Nick has some money in currency S and wonders if he can somehow, after some exchange operations, increase his capital. Of course, he wants to have his money in currency S in the end. Help him to answer this difficult question. Nick must always have non-negative sum of money while making his operations.
Input
The first line of the input contains four numbers: N - the number of currencies, M - the number of exchange points, S - the number of currency Nick has and V - the quantity of currency units he has. The following M lines contain 6 numbers each - the description of the corresponding exchange point - in specified above order. Numbers are separated by one or more spaces. 1<=S<=N<=100, 1<=M<=100, V is real number, 0<=V<=103.
For each point exchange rates and commissions are real, given with at most two digits after the decimal point, 10-2<=rate<=102, 0<=commission<=102.
Let us call some sequence of the exchange operations simple if no exchange point is used more than once in this sequence. You may assume that ratio of the numeric values of the sums at the end and at the beginning of any simple sequence of the exchange operations will be less than 104.
Output
If Nick can increase his wealth, output YES, in other case output NO to the output file.
Sample Input
3 2 1 20.0 1 2 1.00 1.00 1.00 1.00 2 3 1.10 1.00 1.10 1.00
Sample Output
YES
Source
Northeastern Europe 2001, Northern Subregion
借鉴kuangbin大大/article/4680056.html
题意:给定M种货币,有N个兑换点,开始手中有的货币种类是S并且数量为V,问能否从起点出发经过货币的兑换后再回到起点时,货币S的数量会增加
用100A兑换B时,汇率为29.75,手续费为0.39,那么最后得到的B为(100 - 0.39) * 29.75 = 2963.39
分析:是否存在正权回路,使得回路上的顶点的权值能够不断进行松弛
采用Bellman ford算法
原本的Bellman ford算法可以判断是否存在负圈,那么这题就是反向运用Bellman ford
如果存在S到其他点的距离能够不断变大时,存在正权回路
初始化d[S]=V 而源点到其他点的距离初始化为无穷小(0)
#include<cstdio> #include<string> #include<iostream> #include<cstring> #include<cmath> #include<stack> #include<queue> #include<map> #include<stdlib.h> #include<algorithm> #define LL __int64 using namespace std; const int MAXN=200+5; int n,m,S,tot; int MON[2][MAXN]; double RC[2][MAXN],d[MAXN],V; bool flag; bool Bellman() { for(int i=1;i<=n;i++) d[i]=0; d[S]=V; for(int i=1;i<n;i++) { bool tg=false; for(int j=1;j<=tot;j++) { int u=MON[0][j]; int v=MON[1][j]; if(d[v]<(d[u]-RC[1][j])*RC[0][j]) { d[v]=(d[u]-RC[1][j])*RC[0][j]; tg=true; } } if(!tg) return false; } for(int i=1;i<=tot;i++) { int u=MON[0][i]; int v=MON[1][i]; if(d[v]<(d[u]-RC[1][i])*RC[0][i]) return true; } return false; } int main() { //freopen("in.txt","r",stdin); while(scanf("%d %d %d %lf",&n,&m,&S,&V)!=EOF) { tot=1; memset(MON,0,sizeof(MON)); memset(RC,0,sizeof(RC)); for(int i=0;i<m;i++) { int a,b; double c,d,e,f; scanf("%d %d %lf %lf %lf %lf",&a,&b,&c,&d,&e,&f); MON[0][tot]=a; MON[1][tot]=b; RC[0][tot]=c; RC[1][tot]=d; tot++; MON[0][tot]=b; MON[1][tot]=a; RC[0][tot]=e; RC[1][tot]=f; tot++; } if(Bellman()) printf("YES\n"); else printf("NO\n"); } return 0; }
View Code
相关文章推荐
- ARP报文格式详解
- C++11 -----一切从lambda说起
- 微信如何快速加好友
- .NET中的动态编译
- 2015第30周六
- prefixfree.js_无前缀脚本
- C++实现禁忌搜索解决TSP问题
- BestCoder 1st Anniversary B.Hidden String DFS
- HDU 2178 猜数字(二分)
- 第三篇重点面向对象与封装
- Java面试题汇总
- 属性与字段的区别与联系
- Android中处理崩溃异常
- 先序序列和后序序列并不能唯一确定二叉树
- REDIS-命令
- HDOJ 5299 Circles Game 圆嵌套+树上SG
- HDU4282 A very hard mathematic problem 快速幂
- poj 3648 Wedding 【2-sat 经典建图 输出一组可行解 好题】 【tarjan求SCC + 缩点 + 拓扑排序 + 染色】
- 一步一步学习网页开发之问题记录一
- 学习Scala第一天 scala开发环境搭建和helloword解析