POJ 2240 Arbitrage (Floyd)

Arbitrage

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12287		Accepted: 5168

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

Ulm Local 1996

题意：汇率增值问题，给出几对货币之间的汇率，计算是否存在一种兑换方法，可以使货币升值（也就是汇率大于1）。

思路：第一道自己用floyd 做出来的题，虽然还是有些不太了解
只要rate[i][i] 大于1，那就能通过兑换的方法赢利。 floyd的应用，
if (rate[i][j] < rate[i][k]*rate[k][j])
 rate[i][j] = rate[i][k]*rate[k][j];

#include<iostream>
#include<cstdio>
#include<map>
#include<cstring>
#include<string>

using namespace std;

const int INF=0x3f3f3f3f;
const int N=40;

int n,m;
double g[N][N];

int main(){

    //freopen("input.txt","r",stdin);

    int cases=0;
    map<string,int> mp;
    int cnt;
    char s1[100],s2[100];
    while(~scanf("%d",&n) && n){
        mp.clear();
        memset(g,0,sizeof(g));
        int i,j,k;
        cnt=0;
        for(i=0;i<n;i++){
            scanf("%s",s1);
            if(!mp[s1])
                mp[s1]=++cnt;
        }
        scanf("%d",&m);
        double p;
        for(i=0;i<m;i++){
            scanf("%s%lf%s",s1,&p,s2);
            g[mp[s1]][mp[s2]]=p;
        }
        for(k=1;k<=n;k++)
            for(i=1;i<=n;i++)
                for(j=1;j<=n;j++)
                    if(g[i][j]<g[i][k]*g[k][j])
                        g[i][j]=g[i][k]*g[k][j];
        if(g[1][1]>1)
            printf("Case %d: Yes\n",++cases);
        else
            printf("Case %d: No\n",++cases);
    }
    return 0;
}