hdu 1217 Arbitrage 2種類のアルゴリズムACコード、Floyd+Bellman-Ford大水題1枚注意は図面があります~~

Arbitrage
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4998    Accepted Submission(s): 2286
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
   
   
   
   

 
この問題の主な意味は1種の貨幣が1周交換した後に、交換したお金はもとのお金より大きくて、YESを出力して、さもなくばNo.
ところで、私は一日中この水の問題をして面白いですか.の
最後に文字列の処理はmapを用い,mapを数字にマッピングし,配列の下付き文字に対応する.
2つのコードはすべてACで、この問題はあまりにも水です.
Bellman-Fordアルゴリズム:

#include <iostream>
#include <string>
#include <cstring>
#include <cstdio>
#include <map>
#define INF 100000000
#define MAX 40 

using namespace std ;


double dis[MAX],graph[MAX][MAX];
bool Bellman_Ford(int n)
{
	for(int i = 0 ; i <= n ; ++i)
	{
		dis[i] = 0.0;
	}
	dis[0] = 1 ;
	for(int i = 0 ; i < n-1 ; ++i)
	{
		for(int j = 0 ; j < n ; ++j)
		{
			for(int k = 0 ; k < n ; ++k)
			{
				if(graph[k][j] > 0.000001)
				{
					if(dis[j] < dis[k]*graph[k][j])
					{
						dis[j] = dis[k]*graph[k][j] ;
					}
				}
			}
		}
	}
	for(int i = 0 ; i < n ; ++i)
	{
		for(int j = 0 ; j < n ; ++j)
		{
			if(dis[i] < dis[j]*graph[j][i])	//         。 
			{
				return true ;
			}
		}
	}
	return false ;
}

int main()
{
	int n , m , c = 1 ; 
	double rate ;
	map<string,int> search;
	string nameA,nameB,temp ;
	while(cin>>n && n)
	{
		for(int i = 0 ; i < n ; ++i)
		{
			cin>>temp ;
			search.insert(pair<string,int>(temp,i));
		}
		cin>>m;
		memset(graph,0,sizeof(graph)) ;
		for(int j = 0 ; j < m ; ++j)
		{
			cin>>nameA>>rate>>nameB;
			int x = search[nameA] , y = search[nameB] ;
			graph[x][y] = rate ;
		}
		if(Bellman_Ford(n))
		{
			printf("Case %d: Yes
",c++);
		}
		else
		{
			printf("Case %d: No
",c++);
		}
		search.clear() ;
	}
	return 0 ;
}

Floydアルゴリズム:

<pre name="code" class="cpp">#include <iostream>
#include <string>
#include <cstring>
#include <cstdio>
#include <map>
#define INF 100000000
#define MAX 40 

using namespace std ;


double graph[MAX][MAX];
bool Floyd(int n)
{
	for(int k = 0 ; k < n ; ++k)
	{
		for(int i = 0 ; i < n ; ++i)
		{
			for(int j = 0 ; j < n ; ++j)
			{
				if(graph[i][j] < graph[i][k]*graph[k][j])
					graph[i][j] = graph[i][k]*graph[k][j] ;
			}
		}
	}
	for(int i = 0 ; i < n ; ++i)
	{
		if(graph[i][i]>1)
		{
			return true ;
		}
	}
	return false ;
}

int main()
{
	int n , m , c = 1 ; 
	double rate ;
	map<string,int> search;
	string nameA,nameB,temp ;
	while(cin>>n && n)
	{
		memset(graph,0,sizeof(graph)) ;
		for(int i = 0 ; i < n ; ++i)
		{
			cin>>temp ;
			search.insert(pair<string,int>(temp,i));
			graph[i][i] = 1 ;
		}
		cin>>m;
		for(int j = 0 ; j < m ; ++j)
		{
			cin>>nameA>>rate>>nameB;
			int x = search[nameA] , y = search[nameB] ;
			graph[x][y] = rate ;
		}
		if(Floyd(n))
		{
			printf("Case %d: Yes
",c++);
		}
		else
		{
			printf("Case %d: No
",c++);
		}
		search.clear() ;
	}
	return 0 ;
}

君と共に励ます.