首页 > 代码库 > [ACM] hdu 1217 Arbitrage (bellman_ford最短路，判断是否有正权回路或Floyed）

[ACM] hdu 1217 Arbitrage (bellman_ford最短路，判断是否有正权回路或Floyed）

2024-07-12 05:04:54   223人阅读

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
University of Ulm Local Contest 1996


解题思路：

方法1：

一种货币经过兑换其它的货币，经过几轮兑换，最终回到该货币，使其价值变大，这就是“套利”。采用bellman_ford算法，不是求对短路，求最大获利，在bellman算法上改动。判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;//改动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])//改动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，求任意两条边之间的最短路，如果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;
}





联系
我们