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
題目大意：給定N種貨幣及它們之間的匯率，問能否經過一系列交換之后讓貨幣升值？
思路:最短路的變形，只需要把權值+變?yōu)?* 就可以，字符串處理要注意。
code：
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;char a[35][50];double b[35][35];int n;int floyd(){ for(int k=1;k<=n;k++) for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) if(b[i][j] < b[i][k]*b[k][j]) b[i][j] = b[i][k] * b[k][j]; for(int i=1;i<=n;i++) if(b[i][i] > 1.0) return 1; return 0;}int search(char ch[]){ for(int i=1;i<=n;i++) if(strcmp(ch,a[i]) == 0) return i;}int main(){ int tot=0; while(~scanf("%d",&n) && n) { memset(b,0,sizeof(b)); for(int i=1;i<=n;i++) scanf("%s",a[i]); int m; scanf("%d",&m); char c1[50],c2[50]; double rate; for(int i=1;i<=m;i++) { scanf("%s %lf %s",c1,&rate,c2); b[search(c1)][search(c2)] = rate; } printf("Case %d: ",++tot); if(floyd()) printf("Yes/n"); else printf("No/n"); } return 0;}新聞熱點
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