POJ2240——Arbitrage(Floyd算法变形)
Arbitrage
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 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
题目大意：
有N种货币，给出了其中一些货币的汇率。eg：(A 2 B) 代表一块A能换两块B，但不代表两块B能换一块A（有向！）
判断是否可以通过货币之间的转换来赚钱。样例一中的1单位USDollar可以通过转换变为1.05单位的USDollar，赚钱了！
解题思路：
Floyd算法的变形，Edge[i][j]=max(Edge[i][j],Edge[i][m]*Edge[m][j])
初始化时将Edge[i][i]=1，Floyd之后通过判断Edge[i][i]是否大于1，来判断是否可以挣钱！
Code：
1 #include<string> 2 #include<iostream> 3 #include<stdio.h> 4 #include<cstring> 5 #define MAXN 100 6 using namespace std; 7 double edge[MAXN+10][MAXN+10]; 8 string name[MAXN+10]; 9 int N; 10 void floyd() 11 { 12 for (int m=1; m<=N; m++) 13 for (int i=1; i<=N; i++) 14 for (int j=1; j<=N; j++) 15 if (edge[i][m]!=INT_MAX&&edge[m][j]!=INT_MAX&& //Floyd的变形公式 16 edge[i][j]<edge[i][m]*edge[m][j]) 17 edge[i][j]=edge[i][m]*edge[m][j]; 18 } 19 int main() 20 { 21 int M,times=0; 22 while (cin>>N) 23 { 24 times++; 25 for (int i=1; i<=N; i++) 26 for (int j=1; j<=N; j++) 27 if (i!=j) edge[i][j]=INT_MIN; 28 else edge[i][j]=1; 29 if (N==0) break; 30 for (int i=1; i<=N; i++) 31 cin>>name[i]; 32 cin>>M; 33 for (int i=1; i<=M; i++) 34 { 35 string tmp2,tmp1; 36 double dis; 37 cin>>tmp1>>dis>>tmp2; 38 int x1,x2; 39 for (x1=1; x1<=N; x1++) //将字符串转换成对应的标号 40 if (name[x1]==tmp1) break; 41 for (x2=1; x2<=N; x2++) 42 if (name[x2]==tmp2) break; 43 edge[x1][x2]=dis; 44 } 45 floyd(); 46 int cnt=0; 47 for (int i=1;i<=N;i++) 48 if (edge[i][i]>1) //判断是否能通过各种转化挣钱 49 cnt++; 50 if (cnt>=1) printf("Case %d: Yes\n",times); 51 else printf("Case %d: No\n",times); 52 } 53 return 0; 54 }
