HDU 1217 Arbitrage

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

Arbitrage

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

Total Submission(s): 6555    Accepted Submission(s): 3044

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

 

Source

University of Ulm Local Contest 1996

 

Recommend

Eddy

思路：与最短路问题的Floyd算法相似，直接上代码吧。

附上AC代码：

#include <bits/stdc++.h>//#pragma comment(linker, "/STACK:102400000, 102400000")using namespace std;const int maxn = 35;double d[maxn][maxn];int n, m;map<string, int> id;void floyd(){for (int k=0; k<n; ++k)for (int i=0; i<n; ++i)for (int j=0; j<n; ++j)d[i][j] = max(d[i][j], d[i][k]*d[k][j]);}int main(){#ifdef LOCALfreopen("input.txt", "r", stdin);freopen("output.txt", "w", stdout);#endifint cas = 0;while (cin >> n && n != 0){id.clear();string t1, t2;for (int i=0; i<n; ++i){cin >> t1;id[t1] = i;}for (int i=0; i<n; ++i)for (int j=0; j<n; ++j){if (i != j)d[i][j] = 0.0;elsed[i][j] = 1.0;}cin >> m;double t;while (m--){cin >> t1 >> t >> t2;d[id[t1]][id[t2]] = t;}floyd();bool ok = false;for (int i=0; i<n; ++i)if (d[i][i] > 1.0){ok = true;break;}++cas;cout << "Case " << cas << ": ";if (ok)cout << "Yes";elsecout << "No";cout << endl;}return 0;}