poj_2240 Arbitrage

Arbitrage

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 10954	Accepted: 4608

题目链接：http://poj.org/problem?id=2240

Description

Arbitrage is the use of discrepancies in currencyexchange rates to transform one unit of a currency into more than one unit ofthe same currency. For example, suppose that 1 US Dollar buys 0.5 Britishpound, 1 British pound buys 10.0 French francs, and 1 French franc buys 0.21 USdollar. Then, by converting currencies, a clever trader can start with 1 USdollar and buy 0.5 * 10.0 * 0.21 = 1.05 US dollars, making a profit of 5percent.

Your job is to write a program that takes a list of currency exchange rates asinput and then determines whether arbitrage is possible or not.

Input

The input will contain one or more test cases. Om thefirst line of each test case there is an integer n (1<=n<=30),representing the number of different currencies. The next n lines each containthe name of one currency. Within a name no spaces will appear. The next linecontains one integer m, representing the length of the table to follow. Thelast m lines each contain the name ci of a source currency, a real number rijwhich represents the exchange rate from ci to cj and a name cj of thedestination currency. Exchanges which do not appear in the table areimpossible.
Test cases are separated from each other by a blank line. Input is terminatedby a value of zero (0) for n.

Output

For each test case, print one line telling whetherarbitrage is possible or not in the format "Case case: Yes"respectively "Case case: No".

Sample Input

USDollar

BritishPound

FrenchFranc

USDollar 0.5 BritishPound

BritishPound 10.0 FrenchFranc

FrenchFranc 0.21 USDollar

USDollar

BritishPound

FrenchFranc

USDollar 0.5 BritishPound

USDollar 4.9 FrenchFranc

BritishPound 10.0 FrenchFranc

BritishPound 1.99 USDollar

FrenchFranc 0.09 BritishPound

FrenchFranc 0.19 USDollar

Sample Output

Case 1: Yes

Case 2: No

Source

Ulm Local 1996

题意：

就是将货币兑换，问这样兑换能否盈利

解题思路：

根据题意给出的字符串建立图，在用flody算法计算任意两个点之间的距离，用乘号不用加号，最后变量一遍图中（i,i）节点，如果有存在大于1的就输出Yes,否则输出No。

代码：

#include <iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#define MAX 40
#define VALUE 0xffffff
using namespace std;
struct word
{int id;char wd[50];
};
word w[MAX];
double g[MAX][MAX];
int n;void flody()
{for(int k=1;k<=n;k++){for(int j=1;j<=n;j++){for(int i=1;i<=n;i++){if(g[i][j]<g[i][k]*g[k][j]){g[i][j]=g[i][k]*g[k][j];}}}}
}int getId(char *wd)
{for(int i=1;i<=n;i++){if(strcmp(wd,w[i].wd)==0)return w[i].id;}return 0;
}int main()
{int i;char a[MAX],b[MAX];float rate;int flag=1;while(true){memset(g,0,sizeof(g));scanf("%d",&n);if(n==0)break;getchar();for(i=1;i<=n;i++){gets(w[i].wd);w[i].id=i;}int count;scanf("%d",&count);while(count--){getchar();cin>>a>>rate>>b;int id1=getId(a);int id2=getId(b);g[id1][id2]=rate;}flody();for(i=1;i<=n;i++){if(g[i][i]>1){printf("Case %d: Yes\n",flag);break;}}if(i==(n+1)){printf("Case %d: No\n",flag);}flag++;}return 0;
}

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