POJ 1860

Currency Exchange

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 18899		Accepted: 6743

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

解释一下基本题意，SB—hypo有s货币，价值为V，他想获得很多其它的货币。身为一个妻管严，该怎么办？╮(╯▽╰)╭，无奈。。。货币交换公式 (B货币) = （A货币总量-佣金）*汇率。

输入N,M,S,V;

以下M行，line 1 代表货币1 到货币2 的汇率、佣金，货币2 到货币1 的汇率、佣金

line 2 代表货币2 到货币3 的汇率、佣金，货币3 到货币2 的汇率、佣金

打印SB-hypo是否能添加收入？YES/NO

PS: 没什么难度。就是贝尔曼的模板，敲上就差点儿相同A了，可是WA了两次，以为是精度的问题，可是看了一下DISCUSS之后，发现dis[]的初始化不正确，应该为0，而不INF。由于这个题是逆向使用贝尔曼算法，找的不是负环。且没有负环。并代表就有正欢，仅仅能说明可能存在最短路，而是能够使之钱数无限添加的正环。

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
const int INF = 1<<20;
const int N = 110;
const int Size = 9999;
using namespace std;
double mapp[N][N];
double dis[N],ANS;
int n,m,s,l;
double num;
struct node{
    int u,v;
    double hui,yong;
}edge[Size];
void init()
{
    l = 0;//若是找负环，dis[]应该初始化INF
    for(int i = 0;i<N;i++) //这里须要注意。由于找的是正环，初始化应该为0，可是不明确为什么不能为负
        dis[i] = 0;
}
int Bellamn()
{
    int flag = 0;
    for(int i = 0;i<n;i++)
    {
        flag = 0;
        for(int j = 0;j<l;j++)
        {     //曾经是 dis[edge[j].v] > dis[edge[j].u] + edge[j].w; 更新最短路 
            if(dis[edge[j].v] < (dis[edge[j].u] - edge[j].yong)* edge[j].hui) // 更新相对最长路
                {
                    dis[edge[j].v] = (dis[edge[j].u] - edge[j].yong)* edge[j].hui;
                    flag = 1;
                }
        }
        if(flag==0)//更新完成跳出
            break;
    }
    for(int j = 0;j<l;j++)
        {
            if(dis[edge[j].v] < (dis[edge[j].u] - edge[j].yong)* edge[j].hui)//推断正环
                {
                   return 1;
                }
        }
        return 0;
}
int main()
{
    int a,b;
    double c,d,cc,dd;
    while(~scanf("%d%d%d%lf",&n,&m,&s,&num))
    {
        init();
        dis[s] = num;
        for(int i = 0;i<m;i++)
        {
            scanf("%d%d%lf%lf%lf%lf",&a,&b,&c,&d,&cc,&dd);
            edge[l].u = a; edge[l].v = b; edge[l].hui = c; edge[l++].yong = d;
            edge[l].u = b; edge[l].v = a; edge[l].hui = cc; edge[l++].yong = dd;
        }
        int st = Bellamn();
        (st==1)?puts("YES"):puts("NO");
    }
    return 0;
}