UVA_104_Arbitrage

最新推荐文章于 2025-12-03 11:29:18 发布

转载最新推荐文章于 2025-12-03 11:29:18 发布 · 234 阅读

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本文介绍了一种使用弗洛伊德算法寻找货币兑换中是否存在套利机会的方法，并提供了详细的C++实现代码。该算法通过迭代计算所有节点间的最短路径来判断是否存在收益大于1%的循环路径。

#include<iostream>    
#include<sstream>    
#include<string>    
#include<vector>    
#include<list>    
#include<set>    
#include<map>    
#include<stack>    
#include<queue>    
#include<algorithm>    
#include<cmath>    
#pragma warning(disable:4996)    
using std::cin;
using std::cout;
using std::endl;
using std::stringstream;
using std::string;
using std::vector;
using std::list;
using std::pair;
using std::set;
using std::multiset;
using std::map;
using std::multimap;
using std::stack;
using std::queue;
void print(int i, int j, int step,const vector<vector<vector<double>>>&path)
{
	if (step == 0)
	{
		cout << i;
		return;
	}
	print(i, path[i][j][step], step - 1,path);
	cout << ' ' << j;
	return;
}
void floyd_warshall(vector<vector<vector<double>>>&coin, vector<vector<vector<double>>>&path)
{
	int step, m;
	int n = coin.size()-1;
	for (step = 2; step <= n; step++)
	{
		for (int k = 1; k <= n; k++)
		{
			for (int i = 1; i <= n; i++)
			{
				for (int j = 1; j <= n; j++)
				{
					if (coin[i][j][step] < coin[i][k][step - 1] * coin[k][j][1])
					{
						coin[i][j][step] = coin[i][k][step - 1] * coin[k][j][1];
						path[i][j][step] = k;
					}
				}
			}
		}
		int i;
		for (i = 1; i <= n; i++)
		{
			if (coin[i][i][step] > 1.01)
			{
				m = i;
				break;
			}
		}
		if (i <= n)
		{
			break;
		}
	}
	if (step > n)
	{
		cout << "no arbitrage sequence exists";
	}
	else
	{
		print(m, m, step,path);
	}
	cout << endl;
}
int main()
{
	freopen("input.txt","r",stdin);    
	//freopen("output.txt","w",stdout); 
	int n;
	while (cin>>n)
	{
		vector<vector<vector<double>>>coin(n+1,vector<vector<double>>(n+1,vector<double>(n+1)));
		vector<vector<vector<double>>>path(n+1, vector<vector<double>>(n+1, vector<double>(n+1)));
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
			{
				if (i == j)
				{
					coin[i][j][1] = 1;
				}
				else
				{
					cin >> coin[i][j][1];
				}
				path[i][j][1] = j;
			}
		floyd_warshall(coin,path);
	}
	return 0;
}