Arbitrage

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

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

Source

University of Ulm Local Contest 1996

题意：已知n种货币，m对互相兑换的汇率，问是否能够通过货币之间的流转盈利。

解析：Floyd算法，不过是求最大路径，而且汇率之间是乘法的关系。最后只需判断是否存在大于1的环即可。

AC代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <map>
#include <string>
using namespace std;
#define INF 123456
#define MAX_V 35
#define MAX_E 1005

double d[MAX_V][MAX_V];
int n, m;

void Floyd(){
    for(int k=1; k<=n; k++)
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
                d[i][j] = max(d[i][j], d[i][k] * d[k][j]);        //乘法，最大值max
}

int main(){
    #ifdef sxk
        freopen("in.txt", "r", stdin);
    #endif //sxk

    int t = 0;
    string s, e;
    double rate;
    map<string, int> a;
    while(scanf("%d", &n)!=EOF && n){
        for(int i=1; i<=n; i++){
            cin>>s;
            a[s] = i;                         //用map将字符串映射为数字
        }
        for(int i=1; i<=n; i++)
        for(int j=1; j<=n; j++) d[i][j] = (i==j) ? 1 : 0;      //注意初始化
        scanf("%d", &m);
        for(int i=1; i<=m; i++){
            cin>>s>>rate>>e;
            d[ a[s] ][ a[e] ] = max(d[ a[s] ][ a[e] ], rate);   //更新d值
        }
        Floyd();
        printf("Case %d: %s\n", ++t, d[1][1] > 1 ? "Yes" : "No");
        a.clear();
    }
    return 0;
}