I - Arbitrage（判断是否有无正环 II）

题目描述

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

题解：不同于我之前写的求正环 I ，这道题随机的选取一个货币种类，然后进行操作，又回到当前种类的时候，判断金币是否增加
因为n比较小，这里我们可以用floyed的算法，如果大于1 则说明金币增加了

代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<map>
using namespace std;
double arr[50][50];
map<string , int> mp;
int n, m, num;
bool flo()
{for(int k = 1; k <= n; k++)for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++)arr[i][j] = max(arr[i][j], arr[i][k] * arr[k][j]);  //表示i 兑换乘 j 时的最大货币数for(int i = 1; i <= n; i++)if(arr[i][i] >1) return true;  //如果大于1 则说明货币增加了return false;
}
int main()
{string s1, s2;double b;int q = 1;while(cin >> n && n){num = 1;mp.clear();for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++)arr[i][j] = i == j ? 1 : 0;  //i 到 i的兑换比率为1for(int i = 0; i < n; i++){cin >> s1;if(!mp[s1]){mp[s1] = num++;  //将名字转化为编号}}cin >> m;int k = 0;for(int i = 0; i < m; i++){cin >> s1 >> b >> s2;arr[mp[s1]][mp[s2]] = b;  // s1 到 s2 的兑换比率为b}printf("Case %d: ", q++);if(flo()){cout << "Yes" <<endl;}else cout << "No" << endl;}return 0;
}