poj2240

1．链接地址

https://vjudge.net/problem/POJ-2240

2．问题描述

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. 

输入样例

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

输出样例

Case 1: Yes
Case 2: No

3．解题思路

求通过兑换货币的方式能不能实现盈利

其实就是一个判环，只要能通过兑换货币增长货币金额，说明存在环可以一直增加

4．算法实现源代码

#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<vector>
#include<map>
 
using namespace std;
int n,m;
double dist[40];  
struct edge 
{
    int st,ed;
    double rt;
    edge(int sst,int eed,double rtt) : st(sst),ed(eed),rt(rtt) {}
    edge() {}
};
 
vector<edge> G;
map<string ,int> mp;
 
bool Bellman_ford(int v)
{
    memset(dist,0,sizeof(dist));
    dist[v] = 1;
    for(int j = 1;j < n;j++) 
    {
        for(int i = 0;i < G.size();i++) 
        {
            int p1 = G[i].st,p2 = G[i].ed;
            if(dist[p2] < dist[p1] * G[i].rt) 
            { 
                dist[p2] = dist[p1] * G[i].rt;
            }
        }
    }
    for(int i = 0;i < G.size();i++)
    {
        int p1 = G[i].st,p2 = G[i].ed;
        if(dist[p2] < dist[p1] * G[i].rt) 
        {
            return true;    
        }
    }
    return false;
}
 
int main()
{
    int cas = 1;
    string s,ss;
    while(~scanf("%d",&n) &&n) 
    {
        for(int i = 0;i < n;i++) 
        {
            cin >> s;
            mp[s] = i; 
        }
        cin >> m;
        string beg,ends;
        double r;
        G.clear();
        for(int i = 0;i < m;i++) 
        {
            cin >> beg >> r >> ends;  
            G.push_back(edge (mp[beg] ,mp[ends] ,r) );
        }
        printf("Case %d: ",cas++);
        for(int i = 0;i < n;i++) 
        { 
            if(Bellman_ford(i)) 
            {  
                cout << "Yes" << endl;
                break;
            }
            else if(i == n - 1)
                cout << "No" <<endl;
        }
    }
    return 0;
}