POJ 2240 Arbitrage（求正环）

 

Arbitrage

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11008		Accepted: 4629

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

Source

Ulm Local 1996

 

 

 

求正环。

分别用了bellman和floyed算法，好像时间都差不多。

 

 

 

/*

POJ 2240

G++ 784K 797ms



*/





#include<stdio.h>

#include<algorithm>

#include<map>

#include<iostream>

#include<string>

#include<string.h>

using namespace std;



const int MAXN=110;

const int MAXE=1100;

double dist[MAXN];

int edge[MAXE][2];

double d[MAXE];

int tol;



bool bellman(int start,int n)

{

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

    dist[start]=1;

    for(int i=1;i<n;i++)

    {

        bool flag=false;

        for(int j=0;j<tol;j++)

        {

            int u=edge[j][0];

            int v=edge[j][1];

            if(dist[v]<dist[u]*d[j])

            {

                flag=true;

                dist[v]=dist[u]*d[j];

            }

        }

        if(!flag)return false;//没有正环

    }



    for(int j=0;j<tol;j++)

      if(dist[edge[j][1]]<dist[edge[j][0]]*d[j])

        return true;

    return false;

}

map<string,int>mp;

int main()

{

    //freopen("in.txt","r",stdin);

    //freopen("out.txt","w",stdout);

    int n;

    string str1,str2;

    int iCase=0;

    int m;

    while(scanf("%d",&n),n)

    {

        iCase++;

        mp.clear();



        for(int i=1;i<=n;i++)

        {

            cin>>str1;

            mp[str1]=i;

        }

        scanf("%d",&m);

        double tmp;

        tol=0;

        while(m--)

        {

            cin>>str1>>tmp>>str2;

            edge[tol][0]=mp[str1];

            edge[tol][1]=mp[str2];

            d[tol++]=tmp;

        }

        printf("Case %d: ",iCase);

        bool flag=false;

        for(int i=1;i<=n;i++)

        {

            if(bellman(i,n))

            {

                flag=true;

                break;

            }

        }

        if(flag)printf("Yes\n");

        else printf("No\n");

    }

    return 0;

}

 

 

 

 

 

 

程序2：floyed

/*

POJ 2240

G++ 784K 797ms



*/





#include<stdio.h>

#include<algorithm>

#include<map>

#include<iostream>

#include<string>

#include<string.h>

using namespace std;



const int MAXN=110;

double dist[MAXN][MAXN];





void floyed(double dis[][MAXN],int n)//结点编号1-n

{

    int i,j,k;

    for(k=1;k<=n;k++)

       for(i=1;i<=n;i++)

         for(j=1;j<=n;j++)

           if(dis[i][j]<dis[i][k]*dis[k][j])

              dis[i][j]=dis[i][k]*dis[k][j];

}

map<string,int>mp;

int main()

{

    //freopen("in.txt","r",stdin);

    //freopen("out.txt","w",stdout);

    int n;

    string str1,str2;

    int iCase=0;

    int m;

    while(scanf("%d",&n),n)

    {

        iCase++;

        mp.clear();

        for(int i=1;i<=n;i++)

        {

            cin>>str1;

            mp[str1]=i;

        }

        for(int i=1;i<=n;i++)

          for(int j=1;j<=n;j++)

          {

              if(i==j)dist[i][j]=1;

              else dist[i][j]=0;

          }

        scanf("%d",&m);

        double tmp;

        while(m--)

        {

            cin>>str1>>tmp>>str2;

            int a=mp[str1];

            int b=mp[str2];

            dist[a][b]=tmp;

        }

        printf("Case %d: ",iCase);

        bool flag=false;

        floyed(dist,n);

        for(int i=1;i<=n;i++)

        {

            if(dist[i][i]>1)

            {

                flag=true;

                break;

            }

        }

        if(flag)printf("Yes\n");

        else printf("No\n");

    }

    return 0;

}