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 解題思路：

方法1：

一種貨幣經過兌換其它的貨幣，經過幾輪兌換，最終回到該貨幣，使其價值變大，這就是“套利”。採用bellman_ford演算法，不是求對短路，求最大獲利，在bellman演算法上改動。判斷是否含有正權迴路。

程式碼：

#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;//改動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])//改動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：

建立圖，鄰接矩陣，求任意兩條邊之間的最短路，如果dis[i][i]>1 的話，說明存在正權迴路。程式碼中使用到了map, 字串到整型編號的對映

程式碼：

#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;
}