poj2240(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), representing the number of differentcurrencies. The next n lines each contain the name of one currency.Within a name no spaces will appear. The next line contains oneinteger m, representing the length of the table to follow. The lastm lines each contain the name ci of a source currency, a realnumber rij which represents the exchange rate from ci to cj and aname cj of the destination currency. Exchanges which do not appearin 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

UlmLocal 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
#include
#include
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].v]=dis[edge[j].u]*edge[j].w;
           }
       }
    }
    for(inti=1;i<=m;i++)
    {
       if(dis[edge[i].v]
       {
           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
#include
#include
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][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;
}