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

      题目大意：求一笔钱经过各种货币之间的兑换，最后兑换回原来的货币是否会增值，增值的话输出Yes,否则输出No.

      先初始化map数组为0，（其中map[i][i] = 1(自身兑换自身汇率为1)），最后用弗洛伊德算法更新map数组，然后遍历map[i][i]的权值是否大于1，如果大于1就说明增值了。

弗洛伊德算法：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<iostream>
#include<string>
#include<algorithm>

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<n;i++)
    {
        if(aa == name[i])
        {
            return i;
        }
    }
}

int floyd()
{
    for(int k=0;k<n;k++)
    {
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(map[i][j] < map[i][k]*map[k][j])
                {
                    map[i][j] = map[i][k]*map[k][j];
                }
            }
        }
    }
    for(int i=0;i<n;i++)
    {
        if(map[i][i] > 1)
        {
            return 1;
        }
    }
    return 0;
}

int main()
{
    int k = 0;
    while(cin >> n)
    {
        if(n == 0)
        {
            break;
        }
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                map[i][j] = 0;
            }
            map[i][i] = 1;
        }
        getchar();
        for(int i=0;i<n;i++)
        {
            cin >> name[i];
        }
        cin >> m;
        string name1,name2;
        double mm;
        getchar();
        for(int i=0;i<m;i++)
        {
            cin >> 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;
}

贝尔曼福特算法：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<string>
#include<algorithm>

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;i++)
    {
        if(aa == name[i])
        {
            return i;
        }
    }
}
int BF()
{
    for(int i=0;i<n;i++)
    {
        num[i] = 0;
    }
    num[0] = 1;
    for(int i=0;i<t;i++)
    {
        int flag = 0;
        for(int j=0;j<t;j++)
        {
            if(num[q[j].y] < num[q[j].x] * q[j].z)
            {
                num[q[j].y] = num[q[j].x] * q[j].z;
            }
        }
    }
    for(int i=0;i<t;i++)
    {
        if(num[q[i].y] < num[q[i].x] * q[i].z)
        {
            return 1;
        }
    }
    return 0;
}

int main()
{
    int k = 0;
    while(cin >> n)
    {
        t = 0;
        if(n == 0)
        {
            break;
        }
        getchar();
        for(int i=0;i<n;i++)
        {
            cin >> name[i];
        }
        string name1,name2;
        double mm;
        cin >> m;
        getchar();
        for(int i=0;i<m;i++)
        {
            cin >> 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;
}