POJ1860Currency Exchange(Bellman + 正权回路)

Currency Exchange
Time Limit: 1000MS   Memory Limit: 30000K
Total Submissions: 23938   Accepted: 8678

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

题意：N种货币，M种兑换关系，拥有S种货币的数量是V，然后M行分别是兑换关系，A,B两个可以互换的货币的种类，AB的汇率，AB的税，BA的汇率，BA的税，问是否通过某种兑换，让S增值

分析：从s出发，看看是否有一条回路，有的话就能通过这条回来不断增值,s就能保证增值
 1 #include <iostream>
 2 #include <cstring>
 3 #include <vector>
 4 #include <cstdio>
 5 #include <algorithm>
 6 using namespace std;
 7 const int INF = 1 << 30;
 8 const int MAX = 100+5;
 9 const double delta = 1e-8;
10 double dist[MAX];
11 int n,m,s;
12 double v;
13 struct point
14 {
15     int a,b;
16     double rat,cost;
17 };
18 vector<point> edge;
19 int zero(double x)
20 {
21     if(x < -delta)
22         return -1;
23     return x > delta;
24 }
25 bool Bellman_Ford(int s)
26 {
27     for(int i = 1; i <= n; i++)
28     {
29         dist[i] = 0;
30     }
31     dist[s] = v;
32     int len = edge.size();
33     for(int i = 1; i < n; i++)
34     {
35         int flag = 0;
36         for(int j = 0; j < len; j++)
37         {
38             int a = edge[j].a;
39             int b = edge[j].b;
40             double rat = edge[j].rat;
41             double cost = edge[j].cost;
42             double temp = (dist[a] - cost) * rat;
43             if(zero(temp - dist[b]) > 0)  //是大于0，一直当非0来算的
44             {
45                 dist[b] = temp;
46                 flag = 1;
47             }
48         }
49         if(flag == 0)
50             break;
51     }
52     for(int j = 0; j < len; j++)
53     {
54         int a = edge[j].a;
55         int b = edge[j].b;
56         double rat = edge[j].rat;
57         double cost = edge[j].cost;
58         double temp = (dist[a] - cost) * rat;
59         if(zero(temp - dist[b]) > 0)
60         {
61             return true;
62         }
63     }
64     return false;
65 }
66 int main()
67 {
68     while(scanf("%d%d%d%lf", &n,&m,&s,&v) != EOF)
69     {
70         int A,B;
71         double Rab,Cab,Rba,Cba;
72         for(int i = 0; i < m; i++)
73         {
74             point temp;
75             scanf("%d%d%lf%lf%lf%lf",&A,&B,&Rab,&Cab,&Rba,&Cba);
76             temp.a = A;
77             temp.b = B;
78             temp.rat = Rab;
79             temp.cost = Cab;
80             edge.push_back(temp);
81             temp.a = B;
82             temp.b = A;
83             temp.cost = Cba;
84             temp.rat = Rba;
85             edge.push_back(temp);
86         }
87         if(Bellman_Ford(s))
88             printf("YES\n");
89         else
90             printf("NO\n");
91     }
92     return 0;
93 }
View Code

 

posted @ 2015-11-25 21:20  zhaop  阅读(179)  评论(0)  编辑  收藏  举报