POJ训练计划1860_Currency Exchange(最短路_BellmanFord算法)

最新推荐文章于 2021-10-18 20:35:06 发布

「已注销」

最新推荐文章于 2021-10-18 20:35:06 发布

阅读量1.3w

点赞数

CC 4.0 BY-SA版权

分类专栏：集训最短路 POJ

本文链接：https://blog.youkuaiyun.com/juncoder/article/details/25097879

POJ 同时被 3 个专栏收录

99 篇文章

订阅专栏

集训

30 篇文章

订阅专栏

最短路

12 篇文章

订阅专栏

探讨如何通过检测货币兑换网络中的正权回路来判断初始资金是否可以通过一系列兑换操作实现增值。采用Bellman-Ford算法遍历所有可能的兑换路径。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Currency Exchange

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 18686		Accepted: 6656

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<=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

解题报告

还是没明白过来，看着题解敲出来的。。。POJ训练计划最短路。。。

题意

有多种汇币，汇币之间可以交换，这需要手续费，当你用100A币交换B币时，A到B的汇率是29.75，手续费是0.39，那么你可以得到(100 - 0.39) * 29.75 = 2963.3975 B币。问s币的金额经过交换最终得到的s币金额数能否增加

货币的交换是可以重复多次的，所以我们需要找出是否存在正权回路，且最后得到的s金额是增加的

找一条正权回路（正权回路：在这一回路上，顶点的权值能不断增加即能一直进行松弛）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>

using namespace std;
struct egdes
{
    int u,v;
    double r,c;
} edge[1010];
int cnt,n,m,s,a,b;
double rab,cab,V,rba,cba;
int BF()
{
    double r,c;
    int i,j,u,v,f=0;
    double dis[1010];
    for(i=1; i<=n; i++)
        dis[i]=0;
    dis[s]=V;
    for(i=0; i<n-1; i++)
    {
        f=0;
        for(j=0; j<cnt; j++)
        {
            u=edge[j].u;
            v=edge[j].v;
            c=edge[j].c;
            r=edge[j].r;
            if(dis[v]<(dis[u]-c)*r)
            {
                dis[v]=(dis[u]-c)*r;
                f=1;
            }
        }
        //if(!f)break;
    }
    for(i=0; i<cnt; i++)
    {
        u=edge[i].u;
        v=edge[i].v;
        c=edge[i].c;
        r=edge[i].r;
        if(dis[v]<(dis[u]-c)*r)
        {
            dis[v]=(dis[u]-c)*r;
            return 1;
        }
    }
    return 0;
}
int main()
{
    cnt=0;
    cin>>n>>m>>s>>V;
    while(m--)
    {
        cin>>a>>b>>rab>>cab>>rba>>cba;
        edge[cnt].u=a;
        edge[cnt].v=b;
        edge[cnt].r=rab;
        edge[cnt++].c=cab;
        edge[cnt].u=b;
        edge[cnt].v=a;
        edge[cnt].r=rba;
        edge[cnt++].c=cba;
    }
    if(BF())cout<<"YES"<<endl;
    else cout<<"NO"<<endl;
    return 0;
}