# POJ 2240 Arbitrage / ZOJ 1092 Arbitrage / HDU 1217 Arbitrage / SPOJ Arbitrage（图论，环）

### 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

Case 1: Yes
Case 2: No

## 代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<map>
#include<string>
#include<queue>
#include<vector>
using namespace std;

const int maxN=40;
const int inf=2147483647;

class Data
{
public:
int v;
double w;
};

int n,m;
vector<Data> E[maxN];
map<string,int> Map;
queue<int> Q;
bool inqueue[maxN];
double Dist[maxN];

bool spfa(int s);

int main()
{
int ti=0;
while (cin>>n)
{
if (n==0)
break;
Map.clear();
for (int i=1;i<=n;i++)
E[i].clear();
for (int i=1;i<=n;i++)
{
string str;
cin>>str;
Map[str]=i;
}
cin>>m;
for (int i=1;i<=m;i++)
{
string str1,str2;
double r;
cin>>str1>>r>>str2;
E[Map[str1]].push_back((Data){Map[str2],r});
}
bool is_ans=0;
for (int i=1;i<=n;i++)
if (spfa(i)==1)
{
is_ans=1;
break;
}
ti++;
if (is_ans==1)
printf("Case %d: Yes\n",ti);
else
printf("Case %d: No\n",ti);
}
return 0;
}

bool spfa(int s)
{
memset(Dist,0,sizeof(Dist));
memset(inqueue,0,sizeof(inqueue));
while (!Q.empty())
Q.pop();
Dist[s]=1;
Q.push(s);
inqueue[s]=1;
do
{
if (Dist[s]>1)
return 1;
int u=Q.front();
Q.pop();
inqueue[u]=0;
for (int i=0;i<E[u].size();i++)
{
int v=E[u][i].v;
if (Dist[v]<Dist[u]*E[u][i].w)
{
Dist[v]=Dist[u]*E[u][i].w;
if (inqueue[v]==0)
{
Q.push(v);
inqueue[v]=1;
}
}
}
}
while (!Q.empty());
return 0;
}

posted @ 2017-07-24 11:25  SYCstudio  阅读(146)  评论(0编辑  收藏