POJ1860 Currency Exchange

Currency Exchange
Time Limit: 1000MS
 
Memory Limit: 30000K
Total Submissions: 19419
 
Accepted: 6926
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
3. 
For each point exchange rates and commissions are real, given with at most two digits after the decimal point, 10
-2<=rate<=10
2, 0<=commission<=10
2. 
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
4. 
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

============ 화려한 분할선 ==============
제목 대의: 닉은 약간의 돈이 있다. 투자를 해서 부수입을 벌고 싶다. S가지 화폐가 있다. 그들 사이에는 N개 은행이 있는데 각각 약간의 수수료와 환전 계수가 필요하다.
지금 네가 마음대로 바꾸든지 닉에게 돈을 벌 수 있느냐고 물어봐라.
사고방식, Bellman-Ford 알고리즘의 사고방식으로 점차적으로 증가하는 고리가 있는지, 복잡도 o(VE)를 판단한다.

#include <iostream>
#include <cmath>
#include <stdio.h>
#include <algorithm>
#include <ctime>
#include <vector>
#include <cstring>
#include <map>
#include <string>
#include <queue>
//#include <fstream>
using namespace std;
#define LL long long
#define ULL unsigned long long
//#define REP(i,n) for(int i=0;i<n;++i)
#define REP(i,a,b) for(int i=a;i<=b;++i)
#define INFLL (1LL)<<62
#define mset(a) memset(a,0,sizeof a)
#define FR(a) freopen(a,"r",stdin)
#define FW(a) freopen(a,"w",stdout)
#define PI 3.141592654
const LL MOD = 1000000007;
const int maxn=111;

struct E
{
	int a[2];
	double r[2],c[2];
	void scan()
	{
		scanf("%d%d%lf%lf%lf%lf",a,a+1,r,c,r+1,c+1);
	}
}edge[maxn];
double v[maxn];
int m,n,s;
double val;
bool bellman()
{
	mset(v);
	v[m]=val;
	REP(ii,1,s){
		double money=v[ii];
		bool flag = 0;
		REP(j,1,n){
			REP(k,0,1)
				if(v[edge[j].a[k^1]]<(v[edge[j].a[k]]-edge[j].c[k])*edge[j].r[k])
				{
					v[edge[j].a[k^1]]=(v[edge[j].a[k]]-edge[j].c[k])*edge[j].r[k];
					flag=1;
				}

		}
		if(!flag)	break;
	}
	REP(j,1,n){
		REP(k,0,1)
			if(v[edge[j].a[k^1]]<(v[edge[j].a[k]]-edge[j].c[k])*edge[j].r[k])
			{
				v[edge[j].a[k^1]]=(v[edge[j].a[k]]-edge[j].c[k])*edge[j].r[k];
				return true;
			}
	}
	return false;
}

int main()
{
	while (cin>>s>>n>>m>>val)
	{
		REP(ii,1,n)
			edge[ii].scan();
		if(bellman())	puts("YES");
		else			puts("NO");
	}
}