HDU 1217 Arbitrage(弗洛伊德算法)

　　　　　　　　　　   Arbitrage

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

　　　　　　　　　　　　　　　　　　 Total Submission(s): 3053    Accepted Submission(s): 1382

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

 

题目链接1：http://acm.hdu.edu.cn/showproblem.php?pid=1217

题目链接2：http://poj.org/problem?id=2240

 

题目大意：给出N种货币，然后给出M种货币之间的兑换率，询问是否存在一种兑换方式使某种货币盈利。

 

题目分析：首先把N种货币映射到一个字符串数组中，然后利用disMap来保存它们之间的兑换率，

　　     然后利用弗洛伊德求出任意两种货币之间的最大兑换率，然后循环判断是否有disMap[i][i]>1.0,若有输出Yes。

 

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

const int MAX_LEN = 50;

double  disMap[MAX_LEN][MAX_LEN];        //记录各货币之间的兑率
char    moneyName[MAX_LEN][MAX_LEN];     //记录各货币的名称
int     totalMark = 0;                   //标识货币名称所在的下标

void InitMoneyRate()
{
    for(int i = 1; i <= MAX_LEN; i++)
        for(int j = 1; j <= MAX_LEN; j++)
        {
            if(i == j)
                disMap[i][j] = 1.0;
            else
                disMap[i][j] = 0.0;
        }
}

int FindMoneyName(char name[])      //查找name所在的下标位置
{
    for(int i = 1; i <= totalMark; i++)
    {
        if(strcmp(moneyName[i], name) == 0)
            return i;
    }

    totalMark++;
    strcpy(moneyName[totalMark], name);
    return totalMark;
}

void floyd(int N)                     //floyd算法求最短距离
{
    for(int k = 1; k <= N; k++)
        for(int i = 1; i <= N; i++)
            for(int j = 1; j <= N; j++)
                if(disMap[i][j] < disMap[i][k] * disMap[k][j])
                    disMap[i][j] = disMap[i][k] * disMap[k][j];
}

int main()
{
    int  N, M, cnt = 0;
    char nameTemp[MAX_LEN];

    while(scanf("%d", &N) && N != 0)
    {
        cnt++;
        getchar();
        InitMoneyRate();

        for(int i = 0; i < N; i++)
        {
            scanf("%s", nameTemp);
            FindMoneyName(nameTemp);
        }

        double rate;
        char strFirst[MAX_LEN];
        char strSecond[MAX_LEN];
        
        scanf("%d", &M);
        getchar();
        while(M--)
        {
            scanf("%s %lf %s", strFirst, &rate, strSecond);
            int sMark = FindMoneyName(strFirst);
            int eMark = FindMoneyName(strSecond);
            disMap[sMark][eMark] = rate;
        }

        floyd(N);

        bool flag = false;
        for(int i = 1; i <= N; i++)
        {
            if(disMap[i][i] > 1.0)
            {
                flag = true;
                break;
            }
        }

        if(flag)
            printf("Case %d: Yes\n", cnt);
        else
            printf("Case %d: No\n",  cnt);
    }
    return 0;
}