POJ 2240Arbitrage（Floyd）

E - Arbitrage

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2240

Appoint description:  System Crawler  (2015-11-24)

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

Sample Output

Case 1: Yes
Case 2: No

 先输入货币的种类，然后接着是每个货币的兑换关系，问是否有一种货币能增值，也就是通过某种兑换关系是g[i][i] >1

初始化傻逼地将g设成了INF,白白贡献了4次WA

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <map>
using namespace std;
const int MAX = 50;
const int INF = 1 << 15;
int n;
double g[MAX][MAX];
map<string,int> m;
void 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][k] != 0 && g[k][j] != 0 && g[i][j] < g[i][k]*g[k][j])
                {
                    g[i][j] = g[i][k]*g[k][j];
                }
            }
        }
    }
}
int main()
{
    int num = 0;
    while(scanf("%d", &n) != EOF && n)
    {
        int t;
        char temp[100],str[100];
        double rat;
        m.clear();
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n; j++)
                g[i][j] = 0;
        }
        for(int i = 1; i <= n; i++)
        {
            scanf("%s", temp);
            m[temp] = i;
        }
        scanf("%d", &t);
        for(int i = 1; i <= t; i++)
        {
            scanf("%s%lf%s",temp,&rat,str);
            g[ m[temp] ][ m[str] ] = rat;
        }
        Floyd();
        int flag = 0;
        for(int i = 1; i <= n; i++)
        {
            if(g[i][i] != 0 && g[i][i] > 1.0)
            {
                flag = 1;
                break;
            }
        }
        printf("Case %d: ",++num);
        if(flag)
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}