Currency Exchange（最短路）

　　　　　　　　　　　　　　　　　　　　　　　　　　　poj—— 1860 Currency Exchange

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 29851		Accepted: 11245

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

题目大意：

题目中主要是说存在货币兑换点，然后现在手里有一种货币，要各种换来换去，最后再换回去的时候看能不能使原本的钱数增多，每一种货币都有对应的汇率，而货币A到货币B的汇率即为1货币A换得得货币B的数量，但兑换点是要收取佣金的，且佣金从源货币中扣除，例如，你想在汇率29.75，佣金为0.39的兑换点把100美元换成卢布，得到的卢布数即为（100-0.39）*29.75 = 2963.3975.

样例解释：

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

多组输入，第一行中N代表有N种货币可以互相兑换，M代表有M个货币兑换点，S代表这个人手中的的货币的编号，V代表这个人手中拥有的货币数量，底下M行

每行六个数，A,B代表可以交换的货币A和B，剩下的实数RAB,CAB,RBA,CBA，代表A到B的汇率，佣金，B到A的汇率，佣金。以某种兑换方式增加原本的钱数，而且必须兑换为原来的货币。

 

思路：

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。

这一道题虽然需要求的是正权回路，但相应的这道题当中我们需要求的是最长路径，因此和Bellman-Ford算法中判断负环是类似的。

因此，我们只需要修改原本的松弛条件，然后先进行n-1轮松弛，最后再进行一次松弛作为检测存不存在正环就可以了。

 

链接地址：（参考博客）　  http://blog.csdn.net/dgghjnjk/article/details/51684154

代码：

#include<queue>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define N 1010
using namespace std;
int n,m,head[N],sum[N],tot,num,u,v;
bool vis[N];
double dis[N],x,y,xx,yy,money;
struct Edge
{
    int u,v,next;
    double x,y;
}edge[N<<];
int add(int u,int v,double x,double y)//注意：传上来的是double类型的，开始时一直传的是int导致wa
{
    tot++;//从1开始用，开始时一直是在后面++，导致一直wa
    edge[tot].u=u;
    edge[tot].v=v;
    edge[tot].x=x;
    edge[tot].y=y;
    edge[tot].next=head[u];
    head[u]=tot;
}
void begin()
{
    memset(head,,sizeof(head));
    memset(sum,,sizeof(sum));
    memset(dis,,sizeof(dis));
    memset(vis,false,sizeof(vis));
    tot=;
}
int spfa(int s)
{
    queue<int>q;
    dis[s]=money;
    vis[s]=true;
    q.push(s);
    while(!q.empty())
    {
        int x=q.front();
        q.pop();vis[x]=false;
        for(int i=head[x];i;i=edge[i].next)
        {
            int v=edge[i].v;
            if(dis[v]<(dis[x]-edge[i].y)*edge[i].x)
            {
                dis[v]=(dis[x]-edge[i].y)*edge[i].x;
                if(!vis[v])
                {
                    vis[v]=true;
                    q.push(v);
                }
                sum[v]++;
                if(sum[v]>n)
                 ;
            }
        }
    }
    ;
}
int main()
{
    while(scanf("%d %d %d %lf",&n,&m,&num,&money)!=EOF)
    {
        begin();
        ;i<=m;i++)
        {
            scanf("%d %d %lf %lf %lf %lf",&u,&v,&x,&y,&xx,&yy);
            add(u,v,x,y);
            add(v,u,xx,yy);
        }
        ) printf("NO\n");
        else printf("YES\n");
    }
    ;
}