Arbitrage 最短路径的变形 flord

Arbitrage

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8307    Accepted Submission(s): 3839

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

题意是：输入一些货币及她们之间的转化，让判断一下，这中间有没有经过别的货币回来变多了；比如第一组，可以假设你有1美元，经过 BritishPound，FrenchFranc 回到 USDollar的时候就变成了1*0.5*10*0.21=1.05美元 当然是变多了；

map<string ,int > s; 将输入的字符串编号，在用floryd 的变形就可以了   最后判断自己到自己的最大乘积是否大于1；

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<string>
#include<map>
#define INF 0x3f3f3f3f
using namespace std;
int m,n;
double ma[110][110];
void floyd()
{
    int i,k,j;
    for(k=1;k<=m;k++)
    {
        for(i=1;i<=m;i++)
        {
            for(j=1;j<=m;j++)
            {
                ma[i][j]=max(ma[i][j],ma[i][k]*ma[k][j]);
            }
        }
    }
}
int main()
{
    string  a,b;
    double q;
    int k=1;
    map<string,int >s;
    while(scanf("%d",&m),m)
    {
        int t;
        s.clear();
        for(int i=1;i<=m;i++)
        {
            cin>>a;
            s[a]=i;
        }
        for(int i=1;i<=m;i++)
            for(int j=1;j<=m;j++)
                ma[i][j]=0;
        cin>>n;
        for(int i=1;i<=n;i++)
        {
            cin>>a>>q>>b;
            ma[s[a]][s[b]]=q;
        }
        floyd();
        int flag=0;
        for(int i=1;i<=m;i++)
        {
            if(ma[i][i]>1)
            {
                flag=1;
            }
        }


        if(flag)
        {
            printf("Case %d: Yes\n",k++);
        }
        else
        {
            printf("Case %d: No\n",k++);
        }
    }
    return 0;
}

人一我百！人十我万！永不放弃~~~怀着自信的心，去追逐梦想。