zoukankan      html  css  js  c++  java
  • Arbitrage(bellman_ford)

    Arbitrage
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 16652   Accepted: 7004

    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

     1 #include<stdio.h>
     2 #include<string.h>
     3 #include<iostream>
     4 using namespace std;
     5 const int M = 40 , inf = 0x3f3f3f3f;
     6 struct Arbitrage
     7 {
     8     int u , v ;
     9     double r ;
    10 }e[M * M];
    11 int n , m ;
    12 char cur[M][20] ;
    13 char a[20] , b[20] ;
    14 double d[M] ;
    15 
    16 void init (char a[20] , char b[20] , int no)
    17 {
    18     for (int i = 1 ; i <= n ; i++) {
    19         if (strcmp (cur[i] , a) == 0)
    20             e[no].u = i ;
    21         if (strcmp (cur[i] , b) == 0)
    22             e[no].v = i ;
    23     }
    24 }
    25 
    26 int bellman_ford (int o)
    27 {
    28     for (int i = 0 ; i <= n ; i++)
    29         d[i] = 0 ;
    30     d[o] = 1.0 ;
    31     double temp = 1.0 ;
    32     bool flag ;
    33     for (int i = 1 ; i <= n ; i++) {
    34         flag = 1 ;
    35         for (int j = 0 ; j < m ; j++) {
    36             if (d[e[j].v] < d[e[j].u] * e[j].r) {
    37                 d[e[j].v] = d[e[j].u] * e[j].r ;
    38                 flag = 0 ;
    39             }
    40             if (d[o] > temp) {
    41                  return true ;
    42             }
    43         }
    44     }
    45     return false ;
    46 }
    47 
    48 int main ()
    49 {
    50     //freopen ("a.txt" , "r" , stdin) ;
    51     int ans = 1 ;
    52     while (~ scanf ("%d" , &n)) {
    53         if (n == 0)
    54             break ;
    55         getchar () ;
    56         for (int i = 1 ; i <= n ; i++) {
    57             gets (cur[i]) ;
    58         }
    59         scanf ("%d" , &m) ;
    60         for (int i = 0 ; i < m ; i++) {
    61             cin >> a >> e[i].r >> b ;
    62             init (a , b , i) ;
    63         }
    64       /*  for (int i = 0 ; i < m ; i++) {
    65             printf ("u = %d , v = %d , r = %.2f
    " , e[i].u , e[i].v , e[i].r) ;
    66         }*/
    67         int i ;
    68         for (i = 1 ; i <= n ; i++) {
    69             if (bellman_ford (i)) {
    70                 printf ("Case %d: Yes
    " , ans++) ;
    71              //   printf ("Arbitrage num : %d
    " , i) ;
    72                 break ;
    73             }
    74             /*printf ("%d team: 
    " , i) ;
    75             for (int j = 1 ; j <= n ; j++)
    76                 printf ("%.2f " ,d[j]) ;
    77             puts ("") ; */
    78         }
    79         if (i == n + 1)
    80             printf ("Case %d: No
    " , ans++) ;
    81     }
    82     return 0 ;
    83 }
    View Code


     

  • 相关阅读:
    vue使用百度编辑器ueditor踩坑记录
    vue项目之webpack打包静态资源路径不准确
    用自己电脑做网站服务器
    telnet测试端口号
    mongodb,redis,mysql的区别和具体应用场景
    移动应用调试之Inspect远程调试
    @vue/cli 3配置文件vue.config.js
    vue+webpack多个项目共用组件动态打包单个项目
    koa/koa2项目搭建
    用Navicat复制数据库到本地(导入.sql文件运行)
  • 原文地址:https://www.cnblogs.com/get-an-AC-everyday/p/4311415.html
Copyright © 2011-2022 走看看