POJ 2240：Arbitrage：folyd最短路算法变形求有向图的盈利环存在

Arbitrage

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 15069 Accepted: 6333

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

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

Ulm Local 1996

将货币的转换率看做路径的花费，使用folyd算法求得两两货币之间的最大值，通过a到b的最大值乘以b到a的最大值大于1，判断存在盈利环。该题目是有向图的盈利环问题，仍然可以使用最短路算法的变形来求解。

#include<stdio.h>#include<stdlib.h>#include<string.h>int n,lu;double a[35][35],d[35][35];char ming[35][105];char ta[105],tb[105];int tta,ttb;int result;void find(){int i,j;for(i=1;i<=n;i++){if(strcmp(ta,ming[i])==0){tta=i;break;}}for(i=1;i<=n;i++)if(strcmp(tb,ming[i])==0){ttb=i;break;}}void init(){int i,j;double tc;for(i=1;i<=n;i++)scanf("%s",ming[i]);scanf("%d",&lu);for(i=1;i<=n;i++)for(j=1;j<=n;j++)d[i][j]=a[i][j]=0;while(lu--){scanf("%s%lf%s",ta,&tc,tb);find();d[tta][ttb]=a[tta][ttb]=tc;}result=0;}void folyd(){int i,j,k;for(k=1;k<=n;k++){for(i=1;i<=n;i++){for(j=1;j<=n;j++){if(d[i][k]&&d[k][j]&&d[i][j]<d[i][k]*d[k][j])d[i][j]=d[i][k]*d[k][j];}}}}void search(){int i,j;for(i=1;i<=n;i++){if(result)break;for(j=1;j<=n;j++)if(d[i][j]*d[j][i]>1){result=1;break;}}}int main(){int i,j;int c=1;while(scanf("%d",&n)!=EOF&&n){init();folyd();search();if(result)printf("Case %d: Yes\n",c);elseprintf("Case %d: No\n",c);c++;}return 0;}