poj-1860
来源:互联网 发布:java empty string 编辑:程序博客网 时间:2024/04/28 01:58
Currency Exchange
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 12341 Accepted: 4183
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
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
----------------------------------------------------------------------------------------------------------------------
bellman_ford算法求正权环。
----------------------------------------------------------------------------------------------------------------------
Code:
program poj_1860;var dis:array[1..100] of double;r,c:array[1..200] of double;x,y:array[1..200] of longint;n,m,i,s:longint;v:double;function bellman_ford:boolean;var i,j:longint; flag:boolean;beginfor i:=1 to n do dis[i]:=0;dis[s]:=v;for i:=1 to n do beginflag:=false;for j:=1 to m shl 1 doif (dis[x[j]]-c[j])*r[j]>dis[y[j]] then beginflag:=true;dis[y[j]]:=(dis[x[j]]-c[j])*r[j];end;if not flag then exit(false);if i=n then exit(true);end;end;beginreadln(n,m,s,v);for i:=1 to m do beginread(x[i],y[i],r[i],c[i],r[i+m],c[i+m]);x[i+m]:=y[i]; y[i+m]:=x[i];end;if bellman_ford thenwriteln(‘YES’)else writeln(‘NO’);end.
- poj 1860
- poj-1860
- POJ 1860
- poj 1860
- poj 1860
- POJ 1860
- POJ 1860
- POJ-1860
- POJ 1860
- POJ 1860
- poj 1860
- poj-1860
- POJ 1860
- poj 1860
- poj 1860
- poj 1860
- POJ-1860
- poj 1860
- 设计模式的概念
- 标准2 维表问题
- 浅谈个人在瀑布流网页的实现中遇到的问题和解决方法
- Android ListView拖动时背景颜色会变成黑色的解决办法
- 黑马程序员-多线程和套接字
- poj-1860
- 线程池类Executors的运用
- int main(int argc,char * argv[]) windows 下的使用
- poj-2240
- WIN7 访问不了WIN2003解决方法
- poj-1125
- 关于 mysql 别名问题
- poj-2253
- 12 很有用的 Chrome 浏览器命令