Arbitrage POJ

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：已知n种货币，以及m种货币汇率，问通过货币转换，能否使财富增加。输入第一行为n种货币，往下n行为货币名称
然后一个数字为m种货币汇率，往下m行为a，汇率，b。a到b的转换。但输入n为0 时结束测试。输出是否能通过货币转换使财富增加

思路：财富增加的话肯定是以同种货币作比较，否则没有意义，因此题意大致可以变成在一个n个顶点的图中能否找到一个正权环，这里的正权环指财富增加，很明显可用Bellman
这里给的顶点时货币的英文名，为了方便简洁，用map 将货币名 与 编号一一对应。

代码：

#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <iostream>
#define mod 998244353
#define eps 1e-6
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;

//用于存放货币的转换
struct node
{
    //x源来的钱的种类，b转换后钱的种类
    int x,y;
    //s表示汇率
    double s;
};
//ve表示有多少个汇率
vector<node> ve;
//mp用于给货币种类编号
map<string,int> mp;
//dis表示起点转换成其他点后的钱数
double dis[35];
//bellman用于判断是否有正环
//n表示有了n个点，v表示起点
bool bellman(int n,int v)
{
    //初始钱数为0；
    for(int i=1;i<=n;i++)
    {
        dis[i]=0;
    }
    //将起点的钱数设为1，为方便计算
    dis[v]=1;
    //n-1次遍历
    for(int i=1;i<n;i++)
    {
        //对每个汇率进行遍历
        for(int j=0;j<ve.size();j++)
        {
            int a=ve[j].x;
            int b=ve[j].y;
            //更新b之间的钱数
            if(dis[b]<dis[a]*ve[j].s)
            {
                dis[b]=dis[a]*ve[j].s;
            }
        }
    }
    //在进行一次遍历，判断是否有正环
    for(int j=0;j<ve.size();j++)
    {
        int a=ve[j].x;
        int b=ve[j].y;
        //如果钱还能增加，则有正环
        if(dis[b]<dis[a]*ve[j].s)
        {
            return true;
        }
    }
    return false;
}
int main()
{
    int n,ans=1;
    while(scanf("%d",&n)&&n!=0)
    {
        string str;
        for(int i=1;i<=n;i++)
        {
            cin>>str;
            //对钱的种类进行标记
            mp[str]=i;
        }
        int m;
        scanf("%d",&m);
        string a,b;
        node no;
        for(int i=0;i<m;i++)
        {
            cin>>a>>no.s>>b;
            no.x=mp[a];
            no.y=mp[b];
            //记录汇率
            ve.push_back(no);
        }
        printf("Case %d: ",ans++);
        //枚举所有的起点
        for(int i=1;i<=n;i++)
        {
            //判断有正环
            if(bellman(n,i))
            {
                printf("Yes
");
                break;
            }//如果到最后一个起点后海没有正环，则表示财富无法增加
            else if(i==n)
            {
                printf("No
");
            }
        }
        //清除STL容器的内存
        ve.clear();
        mp.clear();
    }
}