poj1860--Currency Exchange

最新推荐文章于 2021-10-18 20:35:06 发布

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Currency Exchange

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 16202		Accepted: 5630

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<=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

题目的意思是输入n,m,s,v,分别代表货币的种类数，兑换点的个数，nick现在的有的货币种类和他现在手头的资金。

接着输入两个兑换点的代号，分别代表R_AB, C_AB, R_BA and C_BA ，最后要你得出nick最后能否通过兑换是自己的资金上涨。

解析，其实就是求最小路径的变形，只不过这里需要判断一下是否有环的存在，这里使用bellman算法，不过需要对判断条件做一个改变。

注意这里的数组需要开的大一点，至少到256，否则就RE

#include<iostream>
#include<cmath>
#include<string>
using namespace std;
int max=220;
double dis[500];
int D[300][2];//存储地址
int n,m,s,i,j,tot;
double C[300][2];//存储汇率以及手续费

bool Bellman(int start,int n,double v)
{
	for(i=1;i<=n;i++)
	{
		dis[i]=0;
	}
	dis[start]=v;
	for(i=1;i<n;i++)
	{
		int flag=0;
		for(j=0;j<tot;j++)//进行松弛
		{
			int u=D[j][0];
			int v=D[j][1];
			if(dis[v]<(dis[u]-C[j][1])*C[j][0])
			{
				flag=1;
				dis[v]=(dis[u]-C[j][1])*C[j][0];
			}
		}
		if(!flag)
			return false;如果没有更新 就退出
	}
	for(i=0;i<tot;i++)
	{
		if(dis[D[i][1]]<(dis[D[i][0]]-C[i][1])*C[i][0])
		{
			return true;寻找正环
		}
	}
	return false;

}
int main()
{
	double v;
	int a,b;
	double c,d,e,f;
	cin>>n>>m>>s>>v;

	 tot=0;
	for(i=0;i<m;i++)
	{
		cin>>a>>b>>c>>d>>e>>f;
		D[tot][0]=a;
		D[tot][1]=b;
		C[tot][0]=c;
		C[tot][1]=d;
		tot++;
		D[tot][0]=b;
		D[tot][1]=a;
		C[tot][0]=e;
		C[tot][1]=f;
		tot++;
	}
	if(Bellman(s,n,v))
		cout<<"YES"<<endl;
	else
		cout<<"NO"<<endl;

return 0;
}