URAL 1162. Currency Exchange

Author: 2015-08-21

ACM

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1. 题目

http://acm.timus.ru/problem.aspx?space=1&num=1162

1162. Currency Exchange

Time limit: 1.0 second
Memory limit: 64 MB
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.

Samples

input output 
3 2 1 10.0
1 2 1.0 1.0 1.0 1.0
2 3 1.1 1.0 1.1 1.0

NO

3 2 1 20.0
1 2 1.0 1.0 1.0 1.0
2 3 1.1 1.0 1.1 1.0

YES
Problem Author: Nick Durov
Problem Source: ACM ICPC 2001. Northeastern European Region, Northern Subregion

2. 思路

现有若干种货币和若干个货币兑换点，每个兑换点单独指定汇率，可以在两种货币间进行兑换，每次兑换收取固定手续费。给出本金，求能否通过货币兑换来获得收益。

把各种货币作为图的节点，对于货币A和货币B，如果有兑换点提供A到B之间的兑换，则A到B、B到A各有一条边，每次兑换后的金额作为边的权值，问题变为求有向带权图的正权环。需要注意这里的权值（即兑换后的金额）需要根据兑换前的金额进行计算。

下面的代码中使用Bellman-Ford算法求解。

3. 代码

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>

#define MAX_CURRENCY_NUM 101
#define MAX_EXCHANGE_POINTS_NUM 101

typedef struct ExchangeRecord {
    int from, to;
    double rate, commission;
} Exchange;

void solveE3c_CurrencyExchange();
void inputExchange(Exchange exchange[], int num);
bool bellmanford(int n, int m, Exchange exchange[], int oriCurrencyId, double oriCurrencyNum);

int main() {
    solveE3c_CurrencyExchange();
    return 0;
}

Exchange exchange[MAX_EXCHANGE_POINTS_NUM * 2];
void solveE3c_CurrencyExchange() {
    int n, m, s;
    double v;
    scanf("%d %d %d %lf", &n, &m, &s, &v);

    inputExchange(exchange, m);
    if (bellmanford(n, m, exchange, s, v)) {
        printf("YES\n");
    }
    else {
        printf("NO\n");
    }
}

void inputExchange(Exchange exchange[], int num) {
    for (int i = 0, j = 0; i < num; ++i) {
        scanf("%d %d %lf %lf %lf %lf", &exchange[j].from, &exchange[j].to, &exchange[j].rate, &exchange[j].commission,
            &exchange[j + 1].rate, &exchange[j + 1].commission);
        exchange[j + 1].from = exchange[j].to;
        exchange[j + 1].to = exchange[j].from;
        j += 2;
    }
}

double balance[MAX_CURRENCY_NUM];
bool bellmanford(int n, int m, Exchange exchange[], int oriCurrencyId, double oriCurrencyNum) {
    for (int i = 1; i <= n; ++i)
        balance[i] = 0;
    balance[oriCurrencyId] = oriCurrencyNum;

    double current;
    bool flag;
    for (int i = 1; i < n; ++i) {
        flag = false;
        for (int j = 0; j < m * 2; ++j) {
            current = (balance[exchange[j].from] - exchange[j].commission)*exchange[j].rate;
            if (balance[exchange[j].to] < current) {
                flag = true;
                balance[exchange[j].to] = current;
            }
        }
        if (flag == false)
            return false;
    }

    for (int i = 0; i < m * 2; ++i) {
        current = (balance[exchange[i].from] - exchange[i].commission)*exchange[i].rate;
        if (balance[exchange[i].to] < current)
            return true;
    }
    return false;
}