POJ2240 Arbitrage【判斷正環的存在】

Arbitrage

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 28447		Accepted: 11926

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

問題連結：POJ2240 Arbitrage

問題描述 ：假設：1美元可以換0.5英鎊，1英鎊可以換10法郎，1法郎可以換0.21美元。那麼1美元經過交換1*0.5*10*0.21=1.05。比本金多了，現在給你n中貨幣，m種兌換方法，如果存在上述情況則返回YES,否則返回NO。

解題思路：判斷是否有正環，使用Bellman_ford演算法，只要修改鬆弛條件和初始化就可以了

AC的C++程式：

#include<iostream>
#include<cstring>
#include<string>
#include<map>

using namespace std;

const int N=10000;
double dist[35];
int n,m;
struct Edge{
	int start,end;
	double w;
}edge[N];

//判斷是否有負環 n個結點，m條邊 
bool Bellman_ford()
{
	memset(dist,0,sizeof(dist));
	dist[1]=1;
	//進行n-1次鬆弛
	for(int i=1;i<n;i++)
	  for(int j=0;j<m;j++)
	    if(dist[edge[j].end]<dist[edge[j].start]*edge[j].w) 
	    	dist[edge[j].end]=dist[edge[j].start]*edge[j].w;
	//判斷是否有正環
	for(int j=0;j<m;j++)
	    if(dist[edge[j].end]<dist[edge[j].start]*edge[j].w) 
	      return true;
	return false;
}

int main()
{
	string s,from,to;
	int cnt=0;
	double c;
	map<string,int>id;
	while(cin>>n&&n){
		for(int i=1;i<=n;i++){
			cin>>s;
			id[s]=i;
		}
		cin>>m;
		for(int i=0;i<m;i++){
			cin>>from>>c>>to;
			edge[i].start=id[from],edge[i].end=id[to],edge[i].w=c;
		}
		cout<<"Case "<<++cnt<<": ";
		if(Bellman_ford())
		  cout<<"Yes"<<endl;
		else
		  cout<<"No"<<endl; 
	}
	return 0;
}