Bellman-Ford-POJ-1860-Currency Exchange

Currency Exchange
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 23595 Accepted: 8542
Description

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 of the input 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 to the output file.
Sample Input

3 2 1 20.0
1 2 1.00 1.00 1.00 1.00
2 3 1.10 1.00 1.10 1.00
Sample Output

YES
Source

Northeastern Europe 2001, Northern Subregion

先将题目抽象为图，每种货币就是点，每个兑换点的一种兑换方式就是一条边，每条边(edge)要存入2种货币类型，汇率以及手续费。
题目要求是否可以增加钱币，那么也就是求是否存在一个权值为正的环。
刚好与Bellman-Ford的松弛条件相反，于是基本算法定下来。
用dis数组来存当全部兑换为第i种货币时，能够得到的最大价值dis[i]。那么每次松弛时，条件就是dis[edge[i].b]<(dis[edge[i].a]-edge[i].c)*edge[i].r。
最后再遍历一次所有边，得出是否存有权值为正的环。
如果有，便输出YES。

//
// main.cpp
// 最短路练习-E-Currency Exchange
//
// Created by 袁子涵 on 15/10/10.
// Copyright (c) 2015年 袁子涵. All rights reserved.
//

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <stdlib.h>

using namespace std;

typedef struct edge
{
    int A,B;
    double r,c;
}Edge;

int N,M,S,sum;
double V,dis[105];
Edge edge[205];

bool Bellman_ford()
{
    bool flag=0;
    memset(dis, 0, sizeof(dis));
    dis[S]=V;
    for (int i=0; i<N; i++) {
        flag=0;
        for (int j=0; j<sum; j++) {
            if (dis[edge[j].B]<(dis[edge[j].A]-edge[j].c)*edge[j].r) {
                dis[edge[j].B]=(dis[edge[j].A]-edge[j].c)*edge[j].r;
                flag=true;
            }
        }
        if (!flag) {
            break;
        }
    }
    for (int i=0; i<sum; i++) {
        if (dis[edge[i].B]<(dis[edge[i].A]-edge[i].c)*edge[i].r) {
            return 0;
        }
    }
    return 1;
}

int main(int argc, const char * argv[]) {
    int a,b;
    double ra,rb,ca,cb;
    cin >> N >> M >> S >> V;
    sum=0;
    for (int i=0; i<M; i++) {
        cin >> a >> b >> rb >> cb >> ra >> ca;
        edge[sum].A=a;
        edge[sum].B=b;
        edge[sum].c=cb;
        edge[sum++].r=rb;
        edge[sum].A=b;
        edge[sum].B=a;
        edge[sum].c=ca;
        edge[sum++].r=ra;
    }
    if (Bellman_ford()) {
        cout << "NO" << endl;
    }
    else
        cout << "YES" << endl;
    return 0;
}