POJ2240 Arbitrage(floyd算法的变形运用)

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本文介绍了一种使用 Floyd 算法来判断是否存在货币套利机会的方法。通过构建货币汇率图并寻找环路中的正权重循环，可以判断是否可以通过一系列交易获得利润。

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Arbitrage

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16745		Accepted: 7035

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

解题思路:

还是floyd的运用，只不过求最短路变成求最长路罢了。

代码如下:

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

const int maxn = 50;
map<string,int> c;
double dist[maxn][maxn];
int m,n;
#define inf 100000

bool floyd()
{
    int k,i,j;
    bool flag;
    for(k=1;k<=m;k++)
    {
        for(i=1;i<=m;i++)
        {
            for(j=1;j<=m;j++)
            {
                if(dist[i][j] < dist[i][k] * dist[k][j])  //寻找最大路径
                {
                    dist[i][j] = dist[i][k] * dist[k][j];
                }
            }
        }
    }
    flag = false;
    for(i=1;i<=m;i++)
    {
        if(dist[i][i] > 1)  //只要找到一种可以赚钱的兑换情况就跳出
        {
            flag = true;
            break;
        }
    }
    return flag;
}

int main()
{
    string s;
    string s1,s2;
    double res;
    int i;
    int cnt;
    cnt = 0;
    //freopen("111","r",stdin);
    while(cin>>m && m)
    {
        memset(dist,inf,sizeof(dist));
        for(i=1;i<=m;i++)
        {
            cin>>s;
            c[s] = i;
            dist[i][i] = 1; //自己到自己的兑换率为1
        }
        cin>>n;
        for(i=1;i<=n;i++)
        {
            cin>>s1>>res>>s2;
            dist[c[s1]][c[s2]] = res;
        }
        if(floyd())
            cout<<"Case "<<++cnt<<": "<<"Yes"<<endl;
        else
            cout<<"Case "<<++cnt<<": "<<"No"<<endl;
    }
    return 0;
}