POJ - 2240 Arbitrage 求有無正環

這個題目沒有規定初始位置，只要有路徑成正環就行。用Floyd算法，計算最長路。如果有一個點到自身的長度大於1，說明存在正環，輸出Yes。

因爲點的名字是字符串，所以用map映射一下。而且輸入數據比較多，直接用cin,cout會超時。

#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iomanip>
#include<map>
#define ll long long
#define ld long double
#define INF 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof a)
using namespace std;
int n,m;
double G[32][32];
map<string,int>  f;                                 //建立一個名字和節點的對應
int floyd()
{
    for(int k=1; k<=n; k++)
    {
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=n; j++)
            {
                if(G[i][j]<G[i][k]*G[k][j])
                    G[i][j]=G[i][k]*G[k][j];
            }
            if(G[i][i]>1)                            //發現正環，直接返回1就行
                return 1;
        }
    }
    return 0;
}
int main()
{
    std::ios::sync_with_stdio(false);               //卡輸入輸出
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    int ccase=1;
    while(cin>>n)
    {
        if(n==0)
            break;
        string a,c;
        double b;
        for(int i=1; i<=n; i++)
        {
            cin>>a;
            f[a]=i;
        }
        cin>>m;
        mem(G,0);
        for(int i=1; i<=m; i++)
        {
            cin>>a>>b>>c;
            G[f[a]][f[c]]=b;
        }
        cout<<"Case "<<ccase++<<": ";
        if(floyd())
            cout<<"Yes"<<endl;
        else
            cout<<"No"<<endl;

    }
    return 0;
}

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