HDU 1217:Arbitrage

Arbitrage

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7688    Accepted Submission(s): 3564

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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

3USDollarBritishPoundFrenchFranc3USDollar 0.5 BritishPoundBritishPound 10.0 FrenchFrancFrenchFranc 0.21 USDollar3USDollarBritishPoundFrenchFranc6USDollar 0.5 BritishPoundUSDollar 4.9 FrenchFrancBritishPound 10.0 FrenchFrancBritishPound 1.99 USDollarFrenchFranc 0.09 BritishPoundFrenchFranc 0.19 USDollar0

 

Sample Output

Case 1: YesCase 2: No
题目意思：给出n中货币，和m个货币间的汇率，让求解是否存在一种货币，先将这种货币兑换为其他货币，再从其他货币兑换回自己。所获得的钱更多。这样就可以通过货币兑换赚取利润。如果存在这样的货币就输出Yes,如果怎样兑换都会赔本，就输出No.
解题思路：
做最短路径专题的时候做到这个题目，但是感觉这样分的真的不好，这个题目根本不关最短路径这个名称什么事，感觉这个分类对我的思考特别的有迷惑性。事实上这个题目可以用最短路径算法的思想去解决。要解决的问题，如何将字符串对应成整型的节点下标？ 可以采用stl的map实现字符串与下标的一一映射。
先开辟一个二维数组，Map[][],Map[i][j]存放i可以兑换成多少j。因此如果存在一种货币k，货币i兑换成货币k,再从货币k兑换成货币j,所得到j货币的数量比从货币i直接兑换成货币j的大。则要更新Map[i][j]要的是i兑换成j的最大j数。其思想与求最短路Flody算法的思想是相同的。所以可以借助Flody算法（这种思想）的模板去解决这个问题。
这个题目和杭电的1596题目find the safest road是相似的。
#include<iostream>#include<cstdio>#include<queue>#include<vector>#include<cstring>#include<string>#include<stack>#include<map>#include<set>#define INF 0x3f3f3f3f#define MAXN 32using namespace std;int n;double Map[MAXN][MAXN];void InitMap()  ///初始化地图{    for(int i = 1; i <= n; i++)    {        Map[i][i] = 1;  ///自身兑换自身的汇率为1        for(int j = i+1; j <= n; j++)        {            Map[i][j] = Map[j][i] = 0;  ///0代表两种货币不能兑换        }    }}void Flody() ///借助Flody求解最短路径的思路求解问题{    for(int k = 1; k <= n; k++)    {        for(int i = 1; i <= n; i++)        {            for(int j = 1; j <= n; j++)            {                if(Map[i][k]*Map[k][j]>Map[i][j]) ///如果货币i兑换成k再兑换成j，比i直接兑换成j所得j货币更多，更新                    Map[i][j] = Map[i][k]*Map[k][j];            }        }    }}int main(){    int t=0;    while(~scanf("%d",&n))    {        if(n == 0) break;        InitMap();        int num;        char name[50],str1[50],str2[50];        double temp;        map<string,int>m;   ///用map来处理货币对应的下标        for(int i = 1; i <= n; i++)        {            scanf("%s",name);            m[name] = i;        }        scanf("%d",&num);        for(int i = 0; i < num; i++)        {            scanf("%s%lf%s",str1,&temp,str2);            int t1 = m[str1];            int t2 = m[str2];            Map[t1][t2] = temp;  ///代表t1可以兑换成temp个t2.        }        Flody();        int flag = 1;        for(int i = 1; i <= n; i++)        {            if(Map[i][i]>1)  ///如果第i种货币经过来回兑换后，比原来钱多了，则就可以通过这样的方式赚到钱            {                printf("Case %d: Yes\n",++t);                flag = 0;                break;            }        }        if(flag) ///代表无论如何兑换，都会赔本，则输出No.            printf("Case %d: No\n",++t);    }    return 0;}