HDOJ1217 floyed算法

Arbitrage

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8575    Accepted Submission(s): 3953

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

3USDollarBritishPoundFrenchFranc3USDollar 0.5 BritishPoundBritishPound 10.0 FrenchFrancFrenchFranc 0.21 USDollar3USDollarBritishPoundFrenchFranc6USDollar 0.5 BritishPoundUSDollar 4.9 FrenchFrancBritishPound 10.0 FrenchFrancBritishPound 1.99 USDollarFrenchFranc 0.09 BritishPoundFrenchFranc 0.19 USDollar0

 

Sample Output

Case 1: YesCase 2: No

 

Source

University of Ulm Local Contest 1996

 

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跟以前的求最短路不同，这里是乘法，而且每次比较取最大值

#include <iostream>#include <cmath>#include <cstring>#include <cstdio>#include <ctime>using namespace std;const int maxn = 35;int n,i,j,k,m,tot;string str[maxn];double a[maxn][maxn];void input(){    string s1,s2;    double tmp;    int p1,p2;    for (i=1; i<=n; i++) cin >> str[i];    for (i=1; i<=n; i++)        for (j=1; j<=n; j++)          a[i][j] = 1;    cin >> m;    for (i=1; i<=m; i++) {        cin >> s1 >> tmp >> s2;        p1 = 1;        p2 = 1;        while (p1<=n && str[p1]!=s1) p1++;        while (p2<=n && str[p2]!=s2) p2++;        a[p1][p2] *= tmp;    }}void floyed(){    for (k=1; k<=n; k++)      for (i=1; i<=n; i++)        for (j=1; j<=n; j++) {            if (a[i][j] < a[i][k]*a[k][j]){                a[i][j] = a[i][k]*a[k][j];            }        }}void output(){    bool flag = 0;    for (i=1; i<=n; i++)        if (a[i][i] > 1) flag = 1;    if (flag) cout << "Case " << ++tot <<": Yes" << endl;    else  cout << "Case " << ++tot <<": No" << endl;}int main(){    std::ios::sync_with_stdio(false);    tot = 0;    while (cin >> n && n) {        input();        floyed();        output();    }    return 0;}