1poj2240(bellman_ford||floyd)

http://poj.org/problem?id=2240

Arbitrage

Time Limit: 1000MSMemory Limit: 65536KTotal Submissions: 11173Accepted: 4704

Description

Arbitrage is the use ofdiscrepancies in currency exchange rates to transform one unit of acurrency into more than one unit of the same currency. For example,suppose that 1 US Dollar buys 0.5 British pound, 1 British poundbuys 10.0 French francs, and 1 French franc buys 0.21 US dollar.Then, by converting currencies, a clever trader can start with 1 USdollar and buy 0.5 * 10.0 * 0.21 = 1.05 US dollars, making a profitof 5 percent.

Your job is to write a program that takes a list of currencyexchange rates as input and then determines whether arbitrage ispossible or not.

Input

The input will contain one ormore test cases. Om the first line of each test case there is aninteger n (1<=n<=30), representingthe number of different currencies. The next n lines each containthe name of one currency. Within a name no spaces will appear. Thenext line contains one integer m, representing the length of thetable to follow. The last m lines each contain the name ci of asource currency, a real number rij which represents the exchangerate 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 isterminated by a value of zero (0) for n.

Output

For each test case, print oneline telling whether arbitrage is possible or not in the format"Case case: Yes" respectively "Case case: No".

Sample Input

3USDollarBritishPoundFrenchFranc3USDollar 0.5 BritishPoundBritishPound 10.0 FrenchFrancFrenchFranc 0.21 USDollar3USDollarBritishPoundFrenchFranc6USDollar 0.5 BritishPoundUSDollar 4.9 FrenchFrancBritishPound 10.0 FrenchFrancBritishPound 1.99 USDollarFrenchFranc 0.09 BritishPoundFrenchFranc 0.19 USDollar0

Sample Output

Case 1: YesCase 2: No

Source

Ulm Local 1996

题意：任意一种货币经过交换后再换回到自己能不能最终得到比原先自己多的钱。

还是bellman_ford()...

这题的数据处理过程值得学习。。。

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
#define maxn 1010
char moneyname[40][40];
struct Edge
{
    int u;
    int v;
    doublew;
}edge[maxn];
double dis[maxn];
int n,m;
int bellman_ford()
{
    for(inti=1;i<=n;i++)
    {
       dis[i]=0;
    }
   dis[1]=1;
    for(inti=1;i<=n-1;i++)
    {
       for(int j=1;j<=m;j++)
       {
           if(dis[edge[j].v]<dis[edge[j].u]*edge[j].w)
           {
               dis[edge[j].v]=dis[edge[j].u]*edge[j].w;
           }
       }
    }
    for(inti=1;i<=m;i++)
    {
       if(dis[edge[i].v]<dis[edge[i].u]*edge[i].w)
       {
           return 1;
       }
    }
    return0;
}
int main()
{
    intcount=1;
   while(scanf("%d",&n)!=EOF)
    {
       if(n==0)
       break;
       for(int i=1;i<=n;i++)
       {
           scanf("%s",moneyname[i]);
       }
       char first[40],last[40];
       double rate;
       int f,l;
       scanf("%d",&m);
       for(int i=1;i<=m;i++)
       {
           scanf("%s%lf%s",first,&rate,last);
           for(int j=1;j<=n;j++)
           {
               if(strcmp(moneyname[j],first)==0)
               {
                   f=j;
               }
               if(strcmp(moneyname[j],last)==0)
               {
                   l=j;
               }
           }
           edge[i].u=f;
           edge[i].v=l;
           edge[i].w=rate;
       }
       printf("Case %d: ",count++);
       if(bellman_ford())
       {
           printf("Yes\n");
       }
       else
       {
           printf("No\n");
       }
    }
    return0;
}

 

代码2：floyd

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define maxn 1010
double edge[maxn][maxn];
char moneyname[40][40];
int n,m;
void floyd()
{
    for(intk=1;k<=n;k++)
    {
       for(int i=1;i<=n;i++)
       {
           for(int j=1;j<=n;j++)
           {
               if(edge[i][j]<edge[i][k]*edge[k][j])
               {
                   edge[i][j]=edge[i][k]*edge[k][j];
               }
           }
       }
    }
}
int main()
{
    intcount=1;
   while(scanf("%d",&n)!=EOF)
    {
       memset(edge,0,sizeof(edge));
       if(n==0)
       {
           break;
       }
       for(int i=1;i<=n;i++)
       {
           scanf("%s",moneyname[i]);
       }
       char first[40],last[40];
       double rate;
       int f,l;
       scanf("%d",&m);
       for(int i=1;i<=m;i++)
       {
           scanf("%s%lf%s",first,&rate,last);
           for(int j=1;j<=n;j++)
           {
               if(strcmp(moneyname[j],first)==0)
               {
                   f=j;
               }
               if(strcmp(moneyname[j],last)==0)
               {
                   l=j;
               }
           }
           edge[f][l]=rate;
       }
       printf("Case %d: ",count++);
       floyd();
       for(int i=1;i<=n;i++)
       {
           if(edge[i][i]>1.0)
           {
               printf("Yes\n");
               break;
           }
           else
           {
               printf("No\n");
               break;
           }
       }
    }
    return0;
}