POJ-2240 Arbitrage(最短路)

Arbitrage

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 23825 Accepted: 10088

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

题意：几种货币中有没有经过几次兑换后使其自身到自身的汇率大于1的，就是任一货币通过其他货币到自身的最大

          最大汇率。

思路：用floyd算法求任意货币到其他货币的最大汇率，判断自身到自身是否大于1。

注意：在初始化时自身到自身应初始化为1，其他初始化为0，虽是Floyd算法但和普通的Floyd还有差别，汇率之间用            乘法而且要求最大路；还有字符串的处理，可选择map容器。

代码：

#include<stdio.h>#include<string>#include<map>#include<iostream>using namespace std;int m,n;double Map[50][50],x;//注意该doule型char u[30],v[30];void Floyd(){    int i,j,k;    for( i=1 ; i<=n ; i++ )        for( j=1 ; j<=n ; j++ )            for( k=1 ; k<=n ; k++ )            if( Map[i][j] < Map[i][k] * Map[k][j] )//最大汇率是小于，汇率之间用乘法                Map[i][j] = Map[i][k] * Map[k][j] ;}int main(){    int q=1;    while(scanf("%d",&n)&&n)    {        map<string,int>mi;//处理字符串        int flag=0;        for( int i=1 ; i<=n ; i++ )        {            for( int j=1 ; j<=n ; j++ )            {                if(i==j)                    Map[i][j]=1;                else                    Map[i][j]=0;            }        }        for( int i=1 ; i<=n ; i++ )//将每个地方标号        {            scanf("%s",u);            mi[u]=i;        }        scanf("%d",&m);        for( int i=1 ; i<=m ; i++ )        {            scanf("%s%lf%s",u,&x,v);            Map[mi[u]][mi[v]]=x;        }        Floyd();        for(int i=1 ; i<=n ; i++ )        {            if(Map[i][i]>1)//汇率增加                flag=1;        }        printf("Case %d: ",q++);        if(flag)            printf("Yes\n");        else            printf("No\n");    }    return 0;}