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
一些价值为s的货币经过一系列的交换，是否能增加价值。抽象出来就是一种货币代表一个节点，节点之间的权值就是交换规则后的价值，因为最后是求最初的货币，因此是一个回路，货币经过这个回路就回升值，说明这个回路是正值的。
运用bellman-ford算法，原始的是判断是否存在负值的环路并求出最短路，这里只要稍微改下松弛操作就可以判断是否存在正值的环路
#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <cstdio>
#include <set>
#include <cmath>
#include <algorithm>
#define INF 0x3f3f3f3f
#define MAXN 105
#define mod 1000000007
using namespace std;
int n,m,s,e;
double v,dis[MAXN];
struct Node
{
int beg,end;
double r,c;
};
Node edge[MAXN<<1];
void add(int b,int en,double r,double c)
{
edge[e].beg=b;
edge[e].end=en;
edge[e].r=r;
edge[e].c=c;
e++;
}
bool relax(int p)
{
double t=(dis[edge[p].beg]-edge[p].c)*edge[p].r;
if(dis[edge[p].end]<t)
{
dis[edge[p].end]=t;
return true;
}
return false;
}
bool bellman()
{
memset(dis,0,sizeof(dis));
dis[s]=v;
for(int i=1; i<n; ++i)
{
bool flag=false;
for(int j=0; j<e; ++j)
{
if(relax(j))
flag=true;
}
if(dis[s]>v)
return true;
if(!flag)
return false;
}
for(int j=0; j<e; ++j)
if(relax(j))
return true;
return false;
}
int main()
{
e=0;
int a,b;
double rab,cab,rba,cba;
scanf("%d%d%d%lf",&n,&m,&s,&v);
for(int i=0; i<m; ++i)
{
scanf("%d%d%lf%lf%lf%lf",&a,&b,&rab,&cab,&rba,&cba);
add(a,b,rab,cab);
add(b,a,rba,cba);
}
if(bellman())
printf("YES\n");
else
printf("NO\n");
return 0;
}