poj 1860 Currency Exchange bellman-ford算法

货币兑换与Bellman-Ford算法

最新推荐文章于 2022-10-10 10:45:18 发布

转载最新推荐文章于 2022-10-10 10:45:18 发布 · 108 阅读

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原文链接：http://www.cnblogs.com/shijianupc/archive/2013/01/24/2875595.html

本文介绍了一个关于货币兑换的问题，通过一系列特定条件下的兑换操作，探讨是否能够增加初始资本。利用Bellman-Ford算法解决该问题，确保在不超过N种货币和M个兑换点的情况下找到是否存在获利机会。

Currency Exchange

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 14644		Accepted: 5050

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 R _AB, C _AB, R _BA and C _BA - 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<=10 ³.
For each point exchange rates and commissions are real, given with at most two digits after the decimal point, 10 ^-2<=rate<=10 ², 0<=commission<=10 ².
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 10 ⁴.

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

Bellman-Ford算法描述：

1,.初始化：将除源点外的所有顶点的最短距离估计值 d[v] ←+∞, d[s] ←0;

2.迭代求解：反复对边集E中的每条边进行松弛操作，使得顶点集V中的每个顶点v的最短距离估计值逐步逼近其最短距离；（运行|v|-1次）

3.检验负权回路：判断边集E中的每一条边的两个端点是否收敛。如果存在未收敛的顶点，则算法返回false，表明问题无解；否则算法返回true，并且从源点可达的顶点v的最短距离保存在 d[v]中。

View Code

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

const int MAX = 105;
int n,m;
int start;
double value;
double dist[MAX];
struct node
{
    int s;
    int t;
    double r;
    double c;
}edge[505];

bool bellman()
{
    int i,j;
    memset(dist, 0, sizeof(dist));
    dist[start] = value;
    for (j=1; j<n; j++)
    {
        bool flag = true;
        for (i=0; i<2*m; i++)
        {
            if (dist[edge[i].t]<(dist[edge[i].s]-edge[i].c)*edge[i].r)
            {
                dist[edge[i].t]=(dist[edge[i].s]-edge[i].c)*edge[i].r;
                flag = false;
            }
        }
        if (flag)break;
    }
    for (i=0; i<2*m; i++)
    {
        if (dist[edge[i].t]<(dist[edge[i].s]-edge[i].c)*edge[i].r)
            return true;
    }
    return false;
}

int main()
{
//    freopen("in.txt","r",stdin);
    int i;
    int a,b;
    while (scanf("%d %d %d %lf",&n,&m,&start,&value)!= EOF)
    {
        for (i=0; i<m; i++)
        {
            scanf("%d %d",&a,&b);
            edge[i].s = a;
            edge[i].t = b;
            scanf("%lf %lf",&edge[i].r,&edge[i].c);
            edge[i+m].s = b;
            edge[i+m].t = a;
            scanf("%lf %lf",&edge[i+m].r,&edge[i+m].c);
        }
        if (bellman())
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}

转载于:https://www.cnblogs.com/shijianupc/archive/2013/01/24/2875595.html