POJ-1860 Currency Exchange（bellman判正环）

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探讨通过一系列货币兑换操作实现资本增值的可能性。面对多种货币及兑换点，分析如何利用不同的汇率和手续费，在不减少初始资金的前提下，最终回到原始货币类型时实现财富的增长。

Currency Exchange

Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 50099 Accepted: 19517
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
Source

“想知道他是否可以在一些交换操作后以某种方式增加他的资本。”

求是否存在正环的问题。

利用最短路算法中的bellman判断负环的性质，稍加修改成判断正环的即可。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
int m,n,s;
double v;
double dis[210];
int cou;
int x1,x2;
double x1h,x1f,x2h,x2f;

struct ch
{
int a,b;
double h,f; 
}sh[210];

void add(int a,int b,double xh,double xf,double yh,double yf)
{
    sh[cou].a=a;
    sh[cou].b=b;
   sh[cou].h=xh;
    sh[cou++].f=xf;
    sh[cou].a=b;
    sh[cou].b=a;
   sh[cou].h=yh;
    sh[cou++].f=yf;
}
int bellman()
{ memset(dis,0,sizeof(dis));
   dis[s]=v;
   for(int i=1;i<=n;i++)
  {
     int f=1;
     for(int j=1;j<cou;j++)
     {
         if(dis[sh[j].b]<(dis[sh[j].a]-sh[j].f)*sh[j].h)
          {
                 f=0;
           dis[sh[j].b]=(dis[sh[j].a]-sh[j].f)*sh[j].h;
          }
      }
       if(f==1)break;
  }
  for(int j=1;j<cou;j++)
{
 if(dis[sh[j].b]<(dis[sh[j].a]-sh[j].f)*sh[j].h)return 1;
}
return 0;
}

int main()
{
while(cin>>n>>m>>s>>v)
{
   cou=1;
   while(m--)
   {
        cin>>x1>>x2>>x1h>>x1f>>x2h>>x2f;
        add(x1,x2,x1h,x1f,x2h,x2f);
   }
   if(bellman())printf("YES\n");
    else printf("NO\n");
}
return 0;
}