POJ---1860 Currency Exchange[套汇问题SPFA()正权回路的判定]

最新推荐文章于 2020-07-23 16:56:37 发布

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最新推荐文章于 2020-07-23 16:56:37 发布

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原文链接：http://www.cnblogs.com/XBWer/archive/2012/08/26/2657722.html

探讨如何通过一系列货币兑换操作实现资金增值的问题。面对多种货币及不同的兑换点，利用图论中的SPFA算法寻找是否存在一条路径使得初始资金能在经过一系列兑换后增加。

Currency Exchange

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 13730		Accepted: 4716

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

Source

Northeastern Europe 2001, Northern Subregion

题意：

直接说测试数据：

3 2 1 20.0

货币的种类 2个兑换点 Nick所拥有的钱的种类拥有货币的数量
1 2 1.00 1.00 1.00 1.00

1 <------> 2 1->2的汇率 1->2的手续费 2->1的汇率 2->1的手续费
2 3 1.10 1.00 1.10 1.00

2 <------> 3 2->3的汇率 2->3的手续费 3->2的汇率 3->2的手续费

code:

 1 #include<iostream>
 2 #include<queue>
 3 using namespace std;
 4 
 5 #define MAXN 110
 6 #define eps 0
 7 int n,m,s;
 8 double v;
 9 double rate[MAXN][MAXN],com[MAXN][MAXN];
10 
11 bool spfa(int st)
12 {
13     queue<int>Que;
14     int cnt[MAXN];
15     bool vst[MAXN];
16     double dis[MAXN];
17     for(int i=0;i<=n;i++)
18     {
19         vst[i]=false;
20         cnt[i]=0;
21         dis[i]=0;
22     }
23     dis[st]=v;
24     vst[st]=1;
25     cnt[st]=1;
26     Que.push(st);
27     while(!Que.empty())
28     {
29         int temp=Que.front();
30         Que.pop();
31         vst[temp]=0;
32         for(int i=1;i<=n;i++)
33         {
34             if(rate[temp][i]>0&&((dis[temp]-com[temp][i])*rate[temp][i])>dis[i])
35             {
36                 dis[i]=(dis[temp]-com[temp][i])*rate[temp][i];
37                 if(!vst[i])
38                 {
39                     vst[i]=true;
40                     Que.push(i);
41                     cnt[i]++;
42                     if(cnt[i]>=n)
43                         return true;
44                 }
45             }
46         }
47     }
48     return dis[s]>v;
49 }
50 
51 int main()
52 {
53     int i;
54     memset(rate,0,sizeof(rate));
55     scanf("%d%d%d%lf",&n,&m,&s,&v);
56     for(i=0;i<m;i++)
57     {
58         int a,b;
59         scanf("%d%d",&a,&b);
60         scanf("%lf%lf%lf%lf",&rate[a][b],&com[a][b],&rate[b][a],&com[b][a]);
61     }
62     if(spfa(s))
63         printf("YES\n");
64     else
65         printf("NO\n");
66     return 0;
67 }