POJ 1860 Currency Exchange

Currency Exchange

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 14358 Accepted: 4946

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 20.01 2 1.00 1.00 1.00 1.002 3 1.10 1.00 1.10 1.00

Sample Output

YES

Source

Northeastern Europe 2001, Northern Subregion

解题思路：Bellman_Ford算法的变型，改变一下初始条件和松弛条件即可，环判断上判断是否有环使得环内任意一点的币种均收敛于一定值，不是的话就证明钱可以通过此环不断循环，以达到增值的的目的。

#include<iostream>using namespace std;int x[10010],y[10010],n,s,edge;double c[10010],r[10010],dis[10010],v;bool Bellman_Ford(){int i,j;bool flag;memset(dis,0,sizeof(dis));//初始化原点到其他点为0，即最小化状态dis[s]=v;//原点到自己距离为现有的钱for(i=2;i<=n;i++)////松弛操作n-1轮，若没有边需要再调整，则退出循环{flag=false;for(j=1;j<=edge;j++)    if((dis[x[j]]-c[j])*r[j]>dis[y[j]])//松弛操作使兑换后利益最大化{dis[y[j]]=(dis[x[j]]-c[j])*r[j];flag=true;    }if(!flag)   break;}for(i=1;i<=edge;i++)if((dis[x[i]]-c[i])*r[i]>dis[y[i]])//若不能使兑换为某一币种的价钱收敛于一定值，则证明在这一个环内反复兑换可使钱币不断增值return true;return false;}int main(){int i,k,S,E;while(cin>>n>>edge>>s>>v){k=0;for(i=1;i<=edge;i++){cin>>S>>E;k++;x[k]=S;y[k]=E;cin>>r[k]>>c[k];k++;x[k]=E;y[k]=S;cin>>r[k]>>c[k];}edge=k;if(Bellman_Ford())cout<<"YES\n";elsecout<<"NO\n";}return 0;}