POJ 2240 Arbitrage

Posted 2020-08-09 风中的簌雨

Arbitrage

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 1   Accepted Submission(s) : 1

Problem 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. <br>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

 

题目大意：可以简单描述为知道从i到j的汇率，问能不能赚钱（就是当i到i初始是t时，从i到i的最长路径能不能大于t，如果有这样的情况就是能赚钱）。

解题思路：该题有多种解法，可以用Bellman-Frod算法判断是否存在是否存在正环，也可以用Floyd算法找出最大环

Bellman-Frod算法：

 1 #include <stdio.h>
 2 #include <string.h>
 3 int n,m,ans;
 4 char str[40][50];
 5 double dis[40];
 6 struct f
 7 {
 8     int x,y;
 9     double z;
10 }rate[1000];
11 int num(char s[])  //输出对应货币的编号
12 {
13     int i;
14     for (i = 1; i <= n; i ++)
15         if (strcmp(str[i],s)==0)
16             break;
17     return i;
18 }
19 void add(int a,int b,double c)
20 {
21     rate[ans].x = a;
22     rate[ans].y = b;
23     rate[ans].z = c;
24     ans ++;
25 }
26 bool Bellman_ford()
27 {
28     int i,j;
29     for (i = 1; i <= n; i ++)
30         dis[i] = 0;
31     dis[1] = 100;
32     for (i = 1; i < n; i ++)
33         for (j = 0; j < ans; j ++)
34             if (dis[rate[j].y] < dis[rate[j].x]*rate[j].z)
35                 dis[rate[j].y] = dis[rate[j].x]*rate[j].z;
36     for (j = 0; j < ans; j ++)
37         if (dis[rate[j].y] < dis[rate[j].x]*rate[j].z)
38             return true;
39     return false;
40 }
41 int main ()
42 {
43     int i,j,f = 1;
44     double c;
45     char ch1[100],ch2[100];
46     while (~scanf("%d",&n))
47     {
48         ans = 0;
49         if (n == 0)
50             break;
51         for (i = 1; i <= n; i ++)
52             scanf(" %s",str[i]);
53 
54 
55         scanf("%d",&m);
56         for (j = 0; j < m; j ++)
57         {
58             scanf("%s %lf %s",ch1,&c,ch2);
59             int a = num(ch1);
60             int b = num(ch2);
61             add(a,b,c);
62         }
63 
64         int flag = 0;
65         printf("Case %d: ",f ++);
66         if (Bellman_ford())
67             printf("Yes\n");
68         else
69             printf("No\n");
70     }
71     return 0;
72 }

View Code

Floyd算法：

 1 //Memory Time   
 2 //276K   79MS   
 3   
 4 #include <iostream>  
 5 #include<map>  
 6 #include<string>  
 7 using namespace std;  
 8   
 9 const int inf=10000;      //无限大  
10 int n;      //货币种类  
11 int m;      //兑换方式  
12   
13 map<string,int>STL;     //建立一个 使字符串与整数有一一对应关系 的容器STL,以便利用邻接矩阵存储数据  
14   
15 double rate;  
16 char str[50],str1[50],str2[50];  
17 double dist[31][31];  
18   
19 int i,j,k;  
20   
21 void floyd(void)  
22 {  
23     for(k=1;k<=n;k++)  
24         for(i=1;i<=n;i++)  
25             for(j=1;j<=n;j++)  
26                 if(dist[i][j] < dist[i][k] * dist[k][j])       //变形的最大路径，变"+"为"*"  
27                     dist[i][j] = dist[i][k] * dist[k][j];  
28     return;  
29 }  
30   
31 int main(void)  
32 {  
33     int cases=1;  
34     while(cases)  
35     {  
36         memset(dist,inf,sizeof(dist));  
37   
38   
39         cin>>n;  
40         if(!n)break;  
41   
42         for(i=1;i<=n;i++)  
43         {  
44             cin>>str;  
45             STL[str]=i;          //将输入的货币从1到n依次编号  
46             dist[i][i]=1;        //到自身的转换率默认为1，但通过floyd可能会被改变  
47                                  //有向图的顶点（一般）存在环  
48         }  
49   
50         cin>>m;  
51         for(i=1;i<=m;i++)  
52         {  
53             cin>>str1>>rate>>str2;  
54             dist[STL[str1]][STL[str2]]=rate;      //构造图  
55         }  
56   
57         floyd();  
58 
59         int flag=false;  
60         for(i=1;i<=n;i++)  
61             if(dist[i][i]>1)  
62             {  
63                 flag=true;  
64                 break;  
65             }  
66         if(flag)  
67             cout<<"Case "<<cases++<<": Yes"<<endl;  
68         else  
69             cout<<"Case "<<cases++<<": No"<<endl;  
70     }  
71     return 0;  
72 }