POJ 2240 Arbitrage

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

首先多组数据输入，每组第一行n，代表接下来有n种货币；然后一个数m， 代表接下来有m组输入， 每组输入为：       货币ci 汇率rate 货币cj，表示货币ci转换为cj的汇率为rate。

很简单的Floyd问题，我们只需要将所有可以互相转换的货币计算一遍，算出转换的汇率，然后最后比较自己与自己转换的汇率是否大于1，大于1则Yes，否则就输出No。  POJ - 1860  这也是一道货币转换题，感觉比这个难一点。

#include <map>
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
double e[35][35];
map<string, int> mat;//存货币 
void init(int n) {
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= n; j++) {
			if(i == j) e[i][j] = 1; //自身的初始汇率是1 
			else e[i][j] = 0;       //初始化为0 
		}
	}
}

void Floyd(int n) {
	for(int k = 1; k <= n; k++) {
		for(int i = 1; i <= n; i++) {
			for(int j = 1; j <= n; j++) {
				if(e[i][j] < (e[i][k]*e[k][j])) {
					e[i][j] = e[i][k]*e[k][j];
				}
			}
		}
	}
}

int main() {
	int n, m;
	int Case = 1;
	char cur[105];//货币 
	double rate;  //汇率 
	char ci[105], cj[105];//货币ci， 货币cj 
	while(~scanf("%d", &n) && n) {
		init(n);
		mat.clear();
		for(int i = 1; i <= n; i++) {
			scanf("%s", cur);
			mat[cur] = i;
		}
		scanf("%d", &m);
		for(int i = 0; i < m; i++) {
			scanf("%s %lf %s", ci, &rate, cj);
			e[mat[ci]][mat[cj]] = rate;
		}
		Floyd(n);
		bool flag = false;
		for(int i = 1; i <= n; i++) {
			if(e[i][i] > 1) {
				flag = true;
				break;
			}
		}
		if(flag) printf("Case %d: Yes\n", Case++);
		else printf("Case %d: No\n",Case++);
	}
	return 0;
}