Currency Exchange (Bellman-Ford算法,判断是否有正权回路)
发布日期:2021-10-16 05:05:06 浏览次数:9 分类:技术文章

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Currency Exchange

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 60000/30000K (Java/Other)
Total Submission(s) : 84   Accepted Submission(s) : 25
Problem 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 R AB, C AB, R BA and C BA - 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<=10<sup>3</sup>.<br>For each point exchange rates and commissions are real, given with at most two digits after the decimal point, 10<sup>-2</sup><=rate<=10<sup>2</sup>, 0<=commission<=10<sup>2</sup>.<br>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 10<sup>4</sup>.<br>
 

Output
If Nick can increase his wealth, output YES, in other case output NO to the output file.
 

Sample Input
3 2 1 20.01 2 1.00 1.00 1.00 1.002 3 1.10 1.00 1.10 1.00
 

Sample Output
YES
 

Source
PKU
 

题意:一个城市有 N 种货币, 有 M 个兑换点, 兑换货币有一定的兑换率 r 和佣金 c,如果货币 A 兑换 B 兑换率是 r ,佣金是 c那么value 个 A 可以换成 (value-c)*r 个 ,  每个兑换点能两种货币双向兑换,但是兑换比例和佣金不同,输入时注意一下, 问:最后如果能够使得自己的钱变多,则输出 YES,否则输出 NO。

思路:

用Bellman-Ford算法判断是否有正权回路,注意是乘不再是加,还有就是应该用">",而不是小于号了。

代码:

#include 
   
    #include 
    
     #include 
     
      #include 
      
       #define inf 0x3f3f3f3fusing namespace std;struct currency{int a,b;double rate,cost;}p[305];int n,m,s;double v,rab,rba,cab,cba,dis[105];int main(){
       
int i,j,x,y;
cin>>n>>m>>s>>v;
for(i=1;i<=m;i++){
cin>>x>>y>>rab>>cab>>rba>>cba;
 //输入
p[i].a=x,p[i].b=y,p[i].rate=rab,p[i].cost=cab;
p[i+m].a=y,p[i+m].b=x,p[i+m].rate=rba,p[i+m].cost=cba;//回来时的,两者是不同的,要注意。
}
for(i=1;i<=n;i++) dis[i]=0;  //初始化,此处与Bellman-Ford的处理相反,初始化为源点到各点距离0,到自身的值为原值
 dis[s]=v;
for(i=1;i<=n;i++)
{
for(j=1;j<=2*m;j++){
if((dis[p[j].a]-p[j].cost)*p[j].rate>dis[p[j].b]) dis[p[j].b]=(dis[p[j].a]-p[j].cost)*p[j].rate;//注意用乘号和大于号 }
}
bool flag=0;
for(i=1;i<=2*m;i++){
if((dis[p[i].a]-p[i].cost)*p[i].rate>dis[p[i].b])
//存在正权回路。
{
flag=1;break;
}
}
if(flag)cout<<"YES"<
else cout<<"NO"<
return 0;}


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很好
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