zoukankan      html  css  js  c++  java
  • [ACM] hdu 1217 Arbitrage (bellman_ford最短路,推断是否有正权回路或Floyed)

    Arbitrage



    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


    解题思路:

    方法1:

    一种货币经过兑换其他的货币,经过几轮兑换。终于回到该货币,使其价值变大,这就是“套利”。採用bellman_ford算法。不是求对短路,求最大获利。在bellman算法上修改。

    推断是否含有正权回路。

    代码:

    #include <iostream>
    #include <string.h>
    using namespace std;
    const int maxn=35;
    double dis[maxn];
    string str[maxn];
    int nodeNum,edgeNum;
    
    struct Edge
    {
        int s,e;
        double w;
    }edge[maxn*maxn];
    
    bool bellman_ford(int start)
    {
        for(int i=1;i<=nodeNum;i++)
            dis[i]=0;//修改1
        dis[start]=1;
        bool ok;
        for(int i=1;i<=nodeNum-1;i++)
        {
            ok=0;
            for(int j=1;j<=edgeNum;j++)
            {
                if(dis[edge[j].s]*edge[j].w>dis[edge[j].e])//修改2
                {
                    dis[edge[j].e]=dis[edge[j].s]*edge[j].w;
                    ok=1;
                }
            }
            if(!ok)
                break;
        }
        for(int i=1;i<=edgeNum;i++)
            if(dis[edge[i].s]*edge[i].w>dis[edge[i].e])
            return true;
        return false;
    }
    
    int main()
    {
        int c=1;
        while(cin>>nodeNum&&nodeNum)
        {
            for(int i=1;i<=nodeNum;i++)
                cin>>str[i];
            cin>>edgeNum;
            string from,to;
            double w;
            for(int i=1;i<=edgeNum;i++)
            {
                cin>>from>>w>>to;
                int j,k;
                for(j=1;j<=nodeNum;j++)
                    if(from==str[j])
                    break;
                for(k=1;k<=nodeNum;k++)
                    if(to==str[k])
                    break;
                edge[i].s=j;
                edge[i].e=k;
                edge[i].w=w;
            }
            if(bellman_ford(1))
                cout<<"Case "<<c++<<": Yes"<<endl;
            else
                cout<<"Case "<<c++<<": No"<<endl;
        }
        return 0;
    }
    
    方法二:

    建立图,邻接矩阵,求随意两条边之间的最短路,假设dis[i][i]>1 的话。说明存在正权回路。代码中使用到了map, 字符串到整型编号的映射

    代码:

    #include <iostream>
    #include <map>
    #include <string.h>
    using namespace std;
    const int maxn=40;
    string str[maxn];
    double mp[maxn][maxn];
    int nodeNum,edgeNum;
    
    void floyed()
    {
        for(int k=1;k<=nodeNum;k++)
            for(int i=1;i<=nodeNum;i++)
                for(int j=1;j<=nodeNum;j++)
                {
                    if(mp[i][j]<mp[i][k]*mp[k][j])
                        mp[i][j]=mp[i][k]*mp[k][j];
                }
    }
    
    int main()
    {
        int c=1;
        while(cin>>nodeNum&&nodeNum)
        {
            map<string,int>st;
            for(int i=1;i<=nodeNum;i++)
            {
                cin>>str[i];
                st[str[i]]=i;
            }
            for(int i=1;i<=nodeNum;i++)
                for(int j=1;j<=nodeNum;j++)
            {
                if(i==j)
                    mp[i][j]=1;
                else
                    mp[i][j]=0;
            }
            cin>>edgeNum;
            string from,to;double w;
            for(int i=1;i<=edgeNum;i++)
            {
                cin>>from>>w>>to;
                mp[st[from]][st[to]]=w;
            }
            floyed();
            bool ok=0;
            for(int i=1;i<=nodeNum;i++)
                if(mp[i][i]>1)
                 ok=1;
            if(ok)
                cout<<"Case "<<c++<<": Yes"<<endl;
            else
                cout<<"Case "<<c++<<": No"<<endl;
        }
        return 0;
    }
    




  • 相关阅读:
    Mysql的联合索引-最左匹配的隐藏规则
    C#读取word文档内容
    安装完office后 在组件服务里DCOM配置中找不到的解决方案
    .NET Web应用程序发布后无法读取Word文档的解决方法
    web程序读取word报异常:COM 类工厂中 CLSID 为 {000209FF-0000-0000-C000-000000000046} 的组件失败,原因是出现以下错误: 80070005 拒绝访问。最新解决方案
    C# 读取txt格式文件内容
    idea 社区版开发 springbook及问题
    Visualvm jvisualvm1.8详情使用
    VSCODE 打造完美java开发环境(新)
    如何将sdk的jar包安装到本地maven库中
  • 原文地址:https://www.cnblogs.com/yfceshi/p/7214668.html
Copyright © 2011-2022 走看看