poj-------(2240)Arbitrage(最短路)

最新推荐文章于 2021-02-04 13:17:31 发布

转载最新推荐文章于 2021-02-04 13:17:31 发布 · 58 阅读

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原文链接：http://www.cnblogs.com/gongxijun/p/3908326.html

本文介绍了一种使用贝尔曼-福特算法检测货币兑换中是否存在盈利机会的方法。通过构建货币兑换网络，利用该算法寻找可能的负权重环，进而判断是否可以通过一系列交易实现利润。

Arbitrage

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15640		Accepted: 6563

Description

Arbitrage is the use of discrepancies in currency exchange rates to transform one unit of a currency into more than one unit of the same currency. For example, suppose that 1 US Dollar buys 0.5 British pound, 1 British pound buys 10.0 French francs, and 1 French franc buys 0.21 US dollar. Then, by converting currencies, a clever trader can start with 1 US dollar and buy 0.5 * 10.0 * 0.21 = 1.05 US dollars, making a profit of 5 percent.

Your job is to write a program that takes a list of currency exchange rates as input and then determines whether arbitrage is possible or not.

Input

The input will contain one or more test cases. Om the first line of each test case there is an integer n (1<=n<=30), representing the number of different currencies. The next n lines each contain the name of one currency. Within a name no spaces will appear. The next line contains one integer m, representing the length of the table to follow. The last m lines each contain the name ci of a source currency, a real number rij which represents the exchange rate from ci to cj and a name cj of the destination currency. Exchanges which do not appear in the table are impossible.
Test cases are separated from each other by a blank line. Input is terminated by a value of zero (0) for n.

Output

For each test case, print one line telling whether arbitrage is possible or not in the format "Case case: Yes" respectively "Case case: No".

Sample Input

3
USDollar
BritishPound
FrenchFranc
3
USDollar 0.5 BritishPound
BritishPound 10.0 FrenchFranc
FrenchFranc 0.21 USDollar

3
USDollar
BritishPound
FrenchFranc
6
USDollar 0.5 BritishPound
USDollar 4.9 FrenchFranc
BritishPound 10.0 FrenchFranc
BritishPound 1.99 USDollar
FrenchFranc 0.09 BritishPound
FrenchFranc 0.19 USDollar

0

Sample Output

Case 1: Yes
Case 2: No

Source

Ulm Local 1996

感觉自己这种算法还是太弱了......唉！没有用floy之间用bellman_floy来做的，bellman算法其实总结起来就是三点：

第一：初始化

第二：循环优化求解最大或者最小

第三：检测是否存在负环

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<string>
 5 #include<iostream>
 6 #include<map>
 7 #include<iterator>
 8 using namespace std;
 9 const double inf =-100;
10 const int maxn = 105;
11 struct node
12 {
13  int u,v;
14  double val;
15 };
16 node edge[1000];
17 double dist[maxn];
18 /*松弛状态判别*/
19 bool relax(int u,int v,double val){
20     if(dist[v]<dist[u]*val){
21         dist[v]=dist[u]*val;
22         return 1;
23     }
24     return 0;
25 }
26 bool Bellman(int st,int n,int m){
27     for(int i=1;i<=n;i++){    //初始化
28         dist[i]=inf;
29     }
30     dist[st]=1;
31     bool flag;
32     /*循环优化部分*/
33     for(int i=1; i<n;i++) {
34           flag=false;
35       for(int j=1;j<=m;j++){
36          if(relax(edge[j].u,edge[j].v,edge[j].val))
37              flag=true;
38          }
39        if(!flag) break;
40     }
41     /*检验部分*/
42     for(int i=1;i<=m;i++){
43        if(relax(edge[i].u,edge[i].v,edge[i].val))
44           return 1;   //有负圈
45     }
46     return 0;
47 }
48 
49 int main()
50 {
51     int n,m;
52     map<string ,int> sac;
53     string temp;
54     int test=1;
55     while(scanf("%d",&n)==1&&n!=0){
56             if(!sac.empty())sac.clear();
57         for(int i=1;i<=n;i++){
58            cin>>temp;
59         // sac.insert(pair<string ,int>(temp,i));
60            sac[temp]=i;
61         }
62          cin>>m;
63          double ss;
64          string aa,bb;
65         map<string,int>::iterator p1,p2;
66         for(int i=1 ; i<=m ; i++ ){
67            cin>>aa>>ss>>bb;
68            p1=sac.find(aa);
69            p2=sac.find(bb);
70            edge[i].u=p1->second;
71            edge[i].v=p2->second;
72            edge[i].val=ss;
73         }
74 
75         if(Bellman(1,n,m)) printf("Case %d: Yes\n",test);
76            else printf("Case %d: No\n",test);
77         test++;
78         // printf("%lf\n",dist[1]);
79     }
80     return 0;
81 }

转载于:https://www.cnblogs.com/gongxijun/p/3908326.html