POJ 2240 Arbitrage(Floyd最短路)

Arbitrage

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 22202 Accepted: 9426

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

题意：给出n种货币，以及m种货币的汇率，问有没有一种货币，经过多次兑换后获得更多的价值（比如1美元经过多次兑换后获得1.2美元），如果有输出Yes,否则输出No;

思路：可以看成一个图，每种货币代表图的顶点，两种货币之间的汇率代表两点之间路径的权值，这样就构成一个有向图，然后用最短路思想改进Floyd算法，算出任意两点间的最大兑换价值，最后查找是否存在一种货币对自己的汇率大于1，，，

AC代码：

#include<stdio.h>
#include<string.h>
char s[105][105],a[105],b[105];
int main(){
    int i,j,k;
    int n,m;
    double v;
    double map[35][35];
    int t=1;
    while(scanf("%d",&n),n){
        for(i=0;i<n;i++){
            for(j=0;j<n;j++){
                if(i==j)
                    map[i][j]=1;  //对自己的汇率为1
                else
                    map[i][j]=0;
            }
        }
        for(i=0;i<n;i++)
            scanf("%s",s[i]);
        scanf("%d",&m);
        for(i=0;i<m;i++){
            scanf("%s%lf%s",a,&v,b);
            for(j=0;j<n;j++){
                if(strcmp(a,s[j])==0)
                    break;
            }
            for(k=0;k<n;k++){
                if(strcmp(b,s[k])==0)
                    break;
            }
            map[j][k]=v;    	//找到对应的汇率并赋值
        }
	///Floyd求任意两点间的最大兑换值
        for(k=0;k<n;k++)
            for(i=0;i<n;i++)
                for(j=0;j<n;j++){
                    if(map[i][k]*map[k][j]>map[i][j])
                        map[i][j]=map[i][k]*map[k][j];
                }


        for(i=0;i<n;i++){	
            if(map[i][i]>1) ///是否存在货币，对自己的汇率大于1
                break;
        }
        if(i==n)
            printf("Case %d: No\n",t);
        else
            printf("Case %d: Yes\n",t);
            t++;
    }
    return 0;
}

POJ 2240 Arbitrage(Floyd最短路)

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