poj2240

robin

Arbitrage

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 20021		Accepted: 8497

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　　题意：求自身到自身的最大转换率。
   最简单的方法就是floryd算法变形，求最大路径后，求最大环，看它是否满足条件。
每一个结点都必须有到自身的环(不甚清楚原因)。
注意：该double的地方一定不能为int，切记!!!



#include<iostream>
#include<cstdio>
#include<map>
#include<string>
#include<cstring>
using namespace std;
const int inf=1e6;      //无限大
int n;//货币种类
int m;//兑换方式
map<string,int>f;     //建立一个 使字符串与整数有一一对应关系 的容器f,以便利用邻接矩阵存储数据
double rate;
char str[50],str1[50],str2[50];
double dist[31][31];
int main()
{
int cases=0;
for(;;)
{
memset(dist,inf,sizeof(dist));
cin>>n;
if(!n) return 0;
for(int i=1;i<=n;i++){
cin>>str;
f[str]=i;          //将输入的货币从1到n依次编号
dist=1;        //到自身的转换率默认为1，但通过floyd可能会被改变     
}//有向图的顶点（一般）存在环
cin>>m;
for(int i=1;i<=m;i++){
cin>>str1>>rate>>str2;
dist[f[str1]][f[str2]]=rate;      //构造图
        }
for(int k=1;k<=n;k++)
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
if(dist[j]<dist[k]*dist[k][j])//变形的最大路径，变"+"为"*"
dist[j]=dist[k]*dist[k][j];
int flag=0;
for(int i=1;i<=n;i++)
if(dist>1){
flag=1;break;
}
if(flag)
cout<<"Case "<<++cases<<": Yes"<<endl;
else
cout<<"Case "<<++cases<<": No"<<endl;
}
return 0;
}