poj 1860 Currency Exchange

Currency Exchange

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 15718		Accepted: 5432

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

我们熟悉的算法dijkstra可用于求单源点到其他各点的最短距离。但是有一个问题在于，图中所有边的权之都应该为正值。因为如果有负权值的边，则只要绕着这条负值边一直走，最短路径很可能越来越短。而这正是bellman ford算法的思想：如果路径可以缩短，那么就按照dijkstra的算法让路径进行N-1次（N为图中的顶点个数）缩短。如果N-1次缩短之后仍能继续缩短，那么图中必定存在负边。
算法的思想参见维基百科里关于bellman Ford算法的解释：
bellman ford算法
       把每个货币币种看作图的一个顶点，把兑换过程看作顶点之间的动态的有向边（注意，是动态的有向边，因为货币数量与兑换完的数量之间是一个函数关系，而不是固定的）。那么：
poj1860这道题，则刚好把“缩短”这个概念换为“货币能增值”，如果经过某个环，让货币增值了，那么就可以让主人公的货币一直增长下去。进行N-1次“增值”之后，如果还能继续增值，则说明图中存在能无限把货币增值的环。
AC代码：

#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <map>
#include <cmath>
#include <vector>
#include <cstdlib>

using namespace std;

struct Edge
{
    int v1,v2;
    double rate,com;
}edges[500];
int n,m,s;
int all;
double v;
double dis[200];
bool bellmanford()
{
    for(int i=0;i<n-1;i++)
    {
        bool flag=false;
        for(int j=0;j<all;j++)
        {
            int v1=edges[j].v1;
            int v2=edges[j].v2;
            if((dis[v1]-edges[j].com)*edges[j].rate>dis[v2])
            {
                dis[v2]=(dis[v1]-edges[j].com)*edges[j].rate;
                flag=true;
            }
        }
        if(!flag) break;
    }
    for(int i=0;i<all;i++)
    {
        int v1=edges[i].v1;
        int v2=edges[i].v2;
        if((dis[v1]-edges[i].com)*edges[i].rate>dis[v2])
        return true;
    }
    return false;
}
int main()
{
  while(cin>>n>>m>>s>>v)
  {
      all=0;
      for(int i=0;i<m;i++)
      {
      cin>>edges[all].v1>>edges[all].v2>>edges[all].rate>>edges[all].com;
      edges[all+1].v1=edges[all].v2;
      edges[all+1].v2=edges[all].v1;
      all++;
      cin>>edges[all].rate>>edges[all].com;
      all++;
      }
      for(int i=1;i<=n;i++)
      dis[i]=0;
      dis[s]=v;
      if(bellmanford())
      cout<<"YES"<<endl;
      else
      cout<<"NO"<<endl;
  }
   return 0;
}