POJ 2240 - Arbitrage

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本文详细阐述了货币套利的概念与操作流程，并通过具体案例演示如何使用Floyd算法来检测是否存在货币套利机会。包括输入输出格式解析、算法实现及运行结果解读。

I - Arbitrage

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2240

Appoint description:

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

用floyd跑一遍，更新条件为g[i][j]<g[i][k]*g[k][j]，初始化g[i][i]=1，则跑完后如果存在g[i][i]>1，则说明存在无限增大的环。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <string>
using namespace std;

#define N 35
int n;
map<string,int> m;
double g[N][N];

bool floyd(){
    for (int k=0;k<n;k++){
        for (int i=0;i<n;i++){
            for (int j=0;j<n;j++){
                if (g[i][j]<g[i][k]*g[k][j])
                    g[i][j]=g[i][k]*g[k][j];
            }
        }
    }

    for (int i=0;i<n;i++)
        if (g[i][i]>1)
            return true;

    return false;
}

int main(){
    int kase=0;
    while (scanf("%d",&n) && n){
        m.clear();
        for (int i=0;i<n;i++){
            string st;
            cin>>st;
            m[st]=i;
        }

        for (int i=0;i<n;i++)
            for (int j=0;j<n;j++)
                g[i][j]=0;
        for (int i=0;i<n;i++)
            g[i][i]=1;

        int t;
        scanf("%d",&t);
        for (int i=0;i<t;i++){
            string st1,st2;
            double d;
            cin>>st1>>d>>st2;
            int u=m[st1],v=m[st2];
            g[u][v]=max(g[u][v],d);
        }

        if (floyd()){
            printf("Case %d: Yes\n",++kase);
        }else{
            printf("Case %d: No\n",++kase);
        }
    }

    return 0;
}