HDU1217 （Floyd简单变形）

Arbitrage

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5794    Accepted Submission(s): 2683

Problem 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 file 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

 

Source

University of Ulm Local Contest 1996

 

套汇问题，floyd变形

 

/*
ID: LinKArftc
PROG: 1217.cpp
LANG: C++
*/

#include <map>
#include <set>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <utility>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-8
#define randin srand((unsigned int)time(NULL))
#define input freopen("input.txt","r",stdin)
#define debug(s) cout << "s = " << s << endl;
#define outstars cout << "*************" << endl;
const double PI = acos(-1.0);
const double e = exp(1.0);
const int inf = 0x3f3f3f3f;
const int INF = 0x7fffffff;
typedef long long ll;

const int maxn = 50;
int n, m;
double table[maxn][maxn];

void floyd() {
    for (int k = 1; k <= n; k ++) {
        for (int i = 1; i <= n; i ++) {
            for (int j = 1; j <= n; j ++) {
                table[i][j] = max(table[i][j], table[i][k] * table[k][j]);
            }
        }
    }
}

int main() {
    //input;
    int _t = 1;
    while (~scanf("%d", &n) && n) {
        map <string, int> mp;
        string str, str1;
        int cnt = 1;
        for (int i = 1; i <= n; i ++) {
            cin >> str;
            mp[str] = cnt ++;
        }
        memset(table, 0, sizeof(table));
        scanf("%d", &m);
        double change;
        for (int i = 1; i <= m; i ++) {
            cin >> str >> change >> str1;
            table[mp[str]][mp[str1]] = change;
        }
        floyd();
        bool flag = false;
        for (int i = 1; i <= n; i ++) {
            if (table[i][i] - 1.0 > eps) {
                flag = true;
                break;
            }
        }
        if (flag) printf("Case %d: Yes
", _t ++);
        else printf("Case %d: No
", _t ++);
    }

    return 0;
}