HDOJ 1217 Arbitrage 【Floyd 有负权】

Arbitrage

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 5804    Accepted Submission(s): 2691

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: YesCase 2: No

题目链接：HDOJ 1217 Arbitrage 【Floyd 有负权】

       货币A 可以购买 X 货币B ，问能否通过一种路径，用X货币能买到比本身更多的货币。
并没有说两种货币可以互相换，可以用Floyd算法变式，将加法换为乘法，
不能交换的两种货币的权值设置为0，只要有一种货币与自己的最短路径大于1就表明能买到比本身更多的货币.

用map映射，将字符串转换为整型编号

已AC代码：

#include<iostream>#include<cstdio>#include<cstring>#include<map>using namespace std;map<string,int> coin;//字符串对整型的映射 double dist[100][100],val;int n,m;char str[100],ch1[100],ch2[100];void Floyd(){int i,j,k;for(k=1;k<=n;++k){for(i=1;i<=n;++i){for(j=1;j<=n;++j){if(dist[i][j]<dist[i][k]*dist[k][j])//加变乘 dist[i][j]=dist[i][k]*dist[k][j];}}}}int main(){int i,j,cnt,CASE=1;while(scanf("%d",&n),n){cnt=1;for(i=0;i<n;++i){scanf("%s",str);coin[str]=cnt++;}memset(dist,0,sizeof(dist));scanf("%d",&m);while(m--){scanf("%s%lf%s",ch1,&val,ch2);dist[coin[ch1]][coin[ch2]]=val;}Floyd();int flag=0;for(i=1;i<=n;++i){if(dist[i][i]>1){flag=1;break;}}if(flag==1)printf("Case %d: Yes\n",CASE++);elseprintf("Case %d: No\n",CASE++);}return 0;}