POJ2240 Arbitrage（最短路）

POJ2240 Arbitrage（最短路）

Arbitrage Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 15991 Accepted: 6737

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

題目大意：求一筆錢經過各種貨幣之間的兌換，最後兌換回原來的貨幣是否會增值，增值的話輸出Yes,否則輸出No.

先初始化map數組為0，（其中map[i][i] = 1(自身兌換自身匯率為1)），最後用弗洛伊德算法更新map數組，然後遍歷map[i][i]的權值是否大於1，如果大於1就說明增值了。

弗洛伊德算法：

#include
#include
#include
#include
#include
#include

using namespace std;
const int N = 10001;
string name[1000];
int n,m;
double map[N][N];

int findd(string aa)
{
    for(int i=0;i 1)
        {
            return 1;
        }
    }
    return 0;
}

int main()
{
    int k = 0;
    while(cin >> n)
    {
        if(n == 0)
        {
            break;
        }
        for(int i=0;i> name[i];
        }
        cin >> m;
        string name1,name2;
        double mm;
        getchar();
        for(int i=0;i> name1 >> mm >> name2;
            int a,b;
            a = findd(name1);
            b = findd(name2);
            map[a][b] = mm;
        }
        int pp = floyd();
        if(pp == 1)
        {
            cout << "Case " << ++k << ": Yes" << endl;
        }
        else
        {
            cout << "Case " << ++k << ": No" << endl;
        }
    }
    return 0;
}

貝爾曼福特算法：

#include
#include
#include
#include
#include

using namespace std;

int n,m;
string name[10000];
int t;
double num[100010];

struct node
{
    int x,y;
    double z;
}q[1000010];

void add(int x,int y,double p)
{
    q[t].x = x;
    q[t].y = y;
    q[t++].z = p;
}

int findd(string aa)
{
    for(int i=0;i> n)
    {
        t = 0;
        if(n == 0)
        {
            break;
        }
        getchar();
        for(int i=0;i> name[i];
        }
        string name1,name2;
        double mm;
        cin >> m;
        getchar();
        for(int i=0;i> name1 >> mm >> name2;
            int a = findd(name1);
            int b = findd(name2);
            add(a,b,mm);
        }
        int pp = BF();
        if(pp == 1)
        {
            cout << "Case " << ++k << ": Yes" << endl;
        }
        else
        {
            cout << "Case " << ++k << ": No" << endl;
        }
    }
    return 0;
}