用c++解决Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operations only with these currencies. There can be several points specializing in the same pair of currencies. Each point has its own exchange rates, exchange rate of A to B is the quantity of B you get for 1A. Also each exchange point has some commission, the sum you have to pay for your exchange operation. Commission is always collected in source currency. For example, if you want to exchange 100 US Dollars into Russian Rubles at the exchange point, where the exchange rate is 29.75, and the commission is 0.39 you will get (100 - 0.39) * 29.75 = 2963.3975RUR. You surely know that there are N different currencies you can deal with in our city. Let us assign unique integer number from 1 to N to each currency. Then each exchange point can be described with 6 numbers: integer A and B - numbers of currencies it exchanges, and real RAB, CAB, RBA and CBA - exchange rates and commissions when exchanging A to B and B to A respectively. Nick has some money in currency S and wonders if he can somehow, after some exchange operations, increase his capital. Of course, he wants to have his money in currency S in the end. Help him to answer this difficult question. Nick must always have non-negative sum of money while making his operations. Input The first line contains four numbers: N - the number of currencies, M - the number of exchange points, S - the number of currency Nick has and V - the quantity of currency units he has. The following M lines contain 6 numbers each - the description of the corresponding exchange point - in specified above order. Numbers are separated by one or more spaces. 1 ≤ S ≤ N ≤ 100, 1 ≤ M ≤ 100, V is real number, 0 ≤ V ≤ 103. For each point exchange rates and commissions are real, given with at most two digits after the decimal point, 10-2 ≤ rate ≤ 102, 0 ≤ commission ≤ 102. Let us call some sequence of the exchange operations simple if no exchange point is used more than once in this sequence. You may assume that ratio of the numeric values of the sums at the end and at the beginning of any simple sequence of the exchange operations will be less than 104. Output If Nick can increase his wealth, output YES, in other case output NO.

这道题目可以用贝尔曼-福德算法（Bellman-Ford Algorithm）来解决。算法的主要思想是通过动态规划的方式来计算从起点到各个顶点的最短路径。在这个问题中，我们可以将起点设置为 Nick 持有的货币，然后通过不断地进行货币兑换来尝试增加他的财富。

具体来说，我们可以定义一个一维数组 d，其中 d[i] 表示从起点到货币 i 的最大收益。初始时，我们将 d[S] 设置为 V，表示 Nick 初始持有的货币数量。然后，我们可以通过遍历所有的兑换点来更新数组 d。对于每个兑换点，我们可以根据其提供的兑换率和佣金计算出从 A 货币到 B 货币的汇率和佣金，然后使用这些值来更新数组 d。具体来说，我们可以使用以下公式来更新 d：

d[B] = max(d[B], (d[A] - C) * R)

其中 C 是佣金，R 是汇率。这个公式的意思是，我们将从 A 货币到 B 货币的汇率应用到当前持有的 A 货币数量上，减去佣金后得到实际能够兑换成的 B 货币数量，然后将这个数量与原本持有的 B 货币数量进行比较，取较大值作为更新后的 d[B] 值。

我们可以重复执行上述更新操作，直到没有任何一次更新能够使 d 数组发生变化为止。如果此时 d[S] 大于 V，那么说明 Nick 可以增加他的财富，输出 YES，否则输出 NO。

下面是完整的 C++ 代码实现：

相关问题

用c++代码实现。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 fran

以下是C++代码实现：

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 210;

int n, m;
double dist[N][N];

int main() {
    while (cin >> n, n) {
        memset(dist, 0, sizeof dist); // 初始化
        for (int i = 1; i <= n; i++) {
            dist[i][i] = 1; // 自己到自己的汇率为1
        }
        cin >> m;
        for (int i = 0; i < m; i++) {
            int a, b;
            double c;
            cin >> a >> b >> c;
            dist[a][b] = c; // 记录汇率
        }
        // Floyd算法求最短路
        for (int k = 1; k <= n; k++) {
            for (int i = 1; i <= n; i++) {
                for (int j = 1; j <= n; j++) {
                    dist[i][j] = max(dist[i][j], dist[i][k] * dist[k][j]);
                }
            }
        }
        bool flag = false;
        for (int i = 1; i <= n; i++) {
            if (dist[i][i] > 1) { // 存在套汇机会
                flag = true;
                break;
            }
        }
        if (flag)
            cout << "Yes" << endl;
        else
            cout << "No" << endl;
    }
    return 0;
}

思路：使用Floyd算法求解任意两个货币之间的汇率，然后检查是否存在套汇机会。若存在，则输出"Yes"，否则输出"No"。

Consider+a+circuit-switched+network.+Suppose+that+all+links+in+the+network+use+TDM+with+10+slots+and

A circuit-switched network is a type of communication network where a dedicated communication path is established between two nodes for the duration of a communication session. In this network, the available bandwidth is divided into fixed-size time slots, and each slot can be used to transmit data or voice signals.

In the given scenario, let's assume that all links in the network use Time Division Multiplexing (TDM) with 10 slots. TDM is a technique that allows multiple signals to be transmitted simultaneously over a single communication channel by dividing the channel into time slots.

With 10 slots in TDM, each link in the circuit-switched network can accommodate up to 10 different communication sessions simultaneously. Each session is allocated a dedicated time slot for transmitting data or voice signals. The time slots are allocated in a cyclic manner, ensuring fair access to all sessions.