zoukankan      html  css  js  c++  java
  • HDOJ1217 Arbitrage[弗洛伊德算法]

    Arbitrage

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 2369    Accepted Submission(s): 1082


    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 file 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
     
    Recommend
    Eddy
     
     
     
     
     
    模版,理解三个点的意义
    code:
      1 #include <iostream>   
      2 #include <iomanip>   
      3 #include <fstream>   
      4 #include <sstream>   
      5 #include <algorithm>   
      6 #include <string>   
      7 #include <set>   
      8 #include <utility>   
      9 #include <queue>   
     10 #include <stack>   
     11 #include <list>   
     12 #include <vector>   
     13 #include <cstdio>   
     14 #include <cstdlib>   
     15 #include <cstring>   
     16 #include <cmath>   
     17 #include <ctime>   
     18 #include <ctype.h> 
     19 using namespace std;
     20 
     21 #define MAXN 35
     22 
     23 char str[MAXN][20];
     24 double map[MAXN][MAXN];
     25 int n,m;
     26 
     27 void init()
     28 {
     29     int i,j;
     30     for(i=0;i<n;i++)
     31         for(j=i;j<n;j++)
     32             map[i][j]=map[j][i]=0;
     33 }
     34 
     35 int fun(char a[])
     36 {
     37     int i;
     38     for(i=0;i<n;i++)
     39         if(strcmp(str[i],a)==0)
     40             return i;
     41 }
     42 
     43 int main()
     44 {
     45     int i,j,k;
     46     char a[20],b[20];
     47     double temp;
     48     int cnt=1;
     49     while(~scanf("%d",&n),n)
     50     {
     51         init();
     52         for(i=0;i<n;i++)
     53             scanf("%s",str[i]);
     54         scanf("%d",&m);
     55         for(i=0;i<m;i++)
     56         {
     57             scanf("%s %lf %s",a,&temp,b);
     58             map[fun(a)][fun(b)]=temp;
     59         }
     60         bool flag=false;
     61         for(k=0;k<n;k++)                                  //中间点
     62         {
     63             for(i=0;i<n;i++)                              //起点
     64             {
     65                 for(j=0;j<n;j++)                           //终点
     66                 {
     67                     if(map[i][j]<map[i][k]*map[k][j])
     68                     {
     69                         map[i][j]=map[i][k]*map[k][j];
     70                         if(map[k][k]>1)
     71                         {
     72                             flag=true;
     73                             break;
     74                         }
     75                     }
     76                 }
     77                 if(flag)
     78                     break;
     79             }
     80             if(flag)
     81                 break;
     82         }
     83         if(flag)
     84             printf("Case %d: Yes\n",cnt++);
     85         else
     86             printf("Case %d: No\n",cnt++);
     87     }
     88     return 0;
     89 }
     90 /*
     91 4
     92 a
     93 b
     94 c
     95 d
     96 9
     97 a 10 b
     98 b 10 c
     99 c 0.2 b
    100 a 0.3 c
    101 c 2 a
    102 c 100 d
    103 d 0.02 c
    104 a 100 d
    105 d 0.001 a
    106 yes
    107 */
  • 相关阅读:
    socket
    RBAC
    CMOS和BIOS
    canvas和SVG
    兼容IE,chrome,ff的设为首页、加入收藏及保存到桌面
    HTML标签marquee实现滚动效果
    百度判断手机终端并自动跳转uaredirect.js代码及使用实例
    JavaScript中常用的事件
    解决windows server 2008 r2 登录进入桌面只显示一片蓝色背景
    ng2自定义管道
  • 原文地址:https://www.cnblogs.com/XBWer/p/2634031.html
Copyright © 2011-2022 走看看