POJ2240 Arbitrage(floyd找正环)

题目:

Arbitrage
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 22203 Accepted: 9427

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

[Submit]   [Go Back]   [Status]   [Discuss]

思路:

先给了一个货币类型n,然后展示了有几种货币,然后给出了各个货币之间的汇率,问从一个货币能不能经过一系列兑换以后钱数增加。用floyd找正环,具体看注释

代码:

#include <cstdio>  
#include <cstring>  
#include <cstdlib>  
#include <cmath>  
#include <string>  
#include <iostream>  
#include <map>  
#include <stack>  
#include <queue>  
#include <vector>  
#include <algorithm>  
#define mem(a,b) memset(a,b,sizeof(a))  
#define N 100+20  
#define M 100000+20  
#define inf 0x3f3f3f3f  
using namespace std;  
map<string,int>mp;  
double mapp[50][50];  
int n,m;  
int main()  
{  
    int q=1;  
    while(~scanf("%d",&n)&&n)  
    {  
        char s1[50],s2[50];  
        mp.clear();  
        for(int i=1; i<=n; i++)  
        {  
            scanf("%s",s1);  
            mp[s1]=i;  
            mapp[i][i]=1;//自己到自己的汇率变成1  
        }  
        scanf("%d",&m);  
        double x;  
        for(int i=1; i<=m; i++)  
        {  
            scanf("%s%lf%s",s1,&x,s2);  
            mapp[mp[s1]][mp[s2]]=x;//用两个键值来赋值  
        }  
        for(int k=1; k<=n; k++)  
            for(int i=1; i<=n; i++)  
                for(int j=1; j<=n; j++)  
                    if(mapp[i][j]<mapp[i][k]*mapp[k][j])  
                        mapp[i][j]=mapp[i][k]*mapp[k][j];  
        bool ok=false;  
        for(int i=1; i<=n; i++)  
        {  
            if(mapp[i][i]>1)//如果自己到自己的汇率增加,证明能获利  
            {  
                ok=true;  
                break;  
            }  
        }  
        if(ok)  
            printf("Case %d: Yes\n",q++);  
        else  
            printf("Case %d: No\n",q++);  
  
    }  
    return 0;  
}  

试一试新博客

点赞

发表评论

电子邮件地址不会被公开。 必填项已用*标注