PKU 百练OJ Arbitrage

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1	http://bailian.openjudge.cn/practice/2240/

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#include <iostream> #include <string> #include <map> using namespace std; std::map<std::string, int> currency; void floyd(float ER[31][31],int n,int cnt) {     for (int k=0;k<n;k++)     {         for (int i=0;i<n;i++)         {             for (int j=0;j<n;j++)             {                 if (ER[i][j]<ER[i][k]*ER[k][j])                 {                     ER[i][j] = ER[i][k] * ER[k][j];                 }             }         }     }       for (int i=0;i<n;i++)     {         if (ER[i][i]>1)         {                           std::cout << "Case " + to_string(cnt) +": Yes" << std::endl;             return;         }     }     std::cout << "Case " + to_string(cnt) + ": No" << std::endl;     return; } int main() {     int n;     int cnt = 1;     while (cin >> n)     {         float ER[31][31] = {0.0};         if (n == 0)         {             return 0;         }         for (int i = 0; i < n; i++)         {             std::string tempString;             cin >> tempString;             currency.emplace(tempString, i);         }         int m;         cin >> m;         for (int i = 0; i < m; i++)         {             std::string tempFirstCurrency, tempSecondCurrency;             float er;             cin >> tempFirstCurrency >> er >> tempSecondCurrency;             ER[currency[tempFirstCurrency]][currency[tempSecondCurrency]] = er;         }         floyd(ER, n, cnt);         cnt++;     }     return 0; }

　　

　　可以参考这篇文章和这个题解

https://www.cnblogs.com/wangyuliang/p/9216365.html

https://www.luogu.org/problemnew/solution/SP9340