hdu1217Arbitrage

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

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: Yes
Case 2: No
这道题用到了两个知识点，一个是map的用法，需要将字符串映射到整数上；另一个就是floyed算法。与普通的floyed算法不同的是，这里将汇率看作边（单向），计算的是任意两个结点
边的乘积的最大值，所以松弛条件略有变化，如果dis【i】【i】的值大于一，则说明存在套利现象，输出Yes，否则输出No。
#include<iostream>#include<cstring>#include<map>using namespace std;#define INF 0x3f3f3f3fdouble dis[31][31];int n,m,cnt=1;void floyed(){int i,j,k,flag=0;for(k=1;k<=n;k++){for(i=1;i<=n;i++){for(j=1;j<=n;j++){if(dis[i][k]<INF&&dis[k][j]<INF){if(dis[i][j]<dis[i][k]*dis[k][j])//注意此处的条件dis[i][j]=dis[i][k]*dis[k][j];}}}}for(i=1;i<=n;i++){if(dis[i][i]>1)flag=1;}if(flag==1)cout<<"Case "<<cnt<<": Yes"<<endl;elsecout<<"Case "<<cnt<<": No"<<endl;}int main(){int i,j;double k;string str[50];string str1,str2;map<string,int >mp;while(cin>>n&&n!=0){memset(dis,0x3f,sizeof(dis));for(i=1;i<=n;i++){cin>>str[i];mp[str[i]]=i;}cin>>m;for(i=0;i<m;i++){cin>>str1>>k>>str2;dis[mp[str1]][mp[str2]]=k;}floyed();cnt++;}}