Hdoj 1217 Arbitrage (Quasi shortest path, Floyd algorithm)

Arbitrage

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 5679 Accepted Submission (s): 2630


Problem Description Arbitrage is the use of discrepancies in currency exchange rates to transform one unit of a currency I Nto 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, MA King a profit of 5 percent.

Your job is to write a program this takes a list of currency exchange rates as input and then determines whether arbitrage is possible or not.

Input the input file would contain one or more test cases. Om the first line of all test case there was an integer n (1<=n<=30), representing the number of different Currencie S. The next n lines each contain the name of one currency. Within a name no spaces would appear. The next line contains one integer m, representing the length of the table to follow. The last m lines all 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 is impossible.
Test cases is separated from 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 are possible or not in the format "case Case:yes" resp ectively "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 Briti Shpound 1.99 USDollar Frenchfranc 0.09 britishpound Frenchfranc 0.19 USDollar 0
Sample Output

Case 1:yes 2:no Test instructions: Arbitrage refers to the use of foreign exchange spreads in different foreign exchange markets, the purchase of a certain currency in a certain foreign exchange market, and the sale of that currency on another foreign exchange market in order to earn profits. This profit is called arbitrage. For example, $1 can be bought for 0.5 pounds, while 1 pounds can buy 10 francs, 1 francs can buy 0.21 dollars, so it is possible to buy 1 dollars by arbitrage using $1.05, arbitrage is there.   The following gives the type and name of each currency, and then gives some currency exchange rate, ask whether there is arbitrage. Solution thinking: can be regarded as the shortest-circuit problem, but the weight of the calculation is multiplied rather than added.   The multiplication of weights with less than 1 leads to a decrease in weight, that is, there is a negative weight problem.     Therefore, this problem can not be used Dijkstra algorithm, the problem may be used Floyd algorithm. The code is as follows:

<span style= "FONT-SIZE:12PX;"
> #include <cstdio> #include <cstring> #define MAXN double MAP[MAXN][MAXN];

int n,t;
	void Floyd () {int i,j,k;
				   for (K=0;k<n;++k) {to (i=0;i<n;++i) {for (j=0;j<n;++j) {if (Map[i][j]<map[i][k]*map[k][j])
			MAP[I][J]=MAP[I][K]*MAP[K][J];
	}}} int sign=0;
			for (I=0;i<n;++i) {if (map[i][i]>1)//its own exchange rate is 1, if the exchange rate is greater than 1 the explanation can arbitrage {sign=1;
		Break
	}} if (sign) printf ("yes\n");
else printf ("no\n");
	} int main () {int m,i,j,a,b,t=1;;
	Double C;
	Char str[32][32],s1[32],s2[32];
		while (scanf ("%d", &n) &&n) {for (i=0;i<n;++i) {for (j=0;j<n;++j) map[i][j]=0;
		} for (I=0;i<n;++i) scanf ("%s", Str[i]);
		scanf ("%d", &m);
			while (m--) {scanf ("%s%lf%s", S1,&AMP;C,S2);
				for (I=0;i<n;++i) {if (strcmp (s1,str[i)) ==0) a=i;
			if (strcmp (S2,str[i]) ==0) b=i;
		} map[a][b]=c;
		} printf ("Case%d:", t++);
	Floyd ();
} return 0; }</SPAN>