poj1860(bellman ford 算法)

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

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

在此我们又学习了一种新的算法——bellman ford 算法。

　　Dijkstra算法中不允许边的权是负权，如果遇到负权，则可以采用Bellman-Ford算法。

　　Bellman-Ford算法能在更普遍的情况下（存在负权边）解决单源点最短路径问题。对于给定的带权（有向或无向）图 G=（V,E），其源点为s，加权函数 w是
边集 E 的映射。对图G运行Bellman-Ford算法的结果是一个布尔值，表明图中是否存在着一个从源点s可达的负权回路。若不存在这样的回路，算法将给出从源点s到
图G的任意顶点v的最短路径d[v]。

　　适用条件&范围

　　1.单源最短路径(从源点s到其它所有顶点v);

　　2.有向图&无向图(无向图可以看作(u,v),(v,u)同属于边集E的有向图);

　　3.边权可正可负(如有负权回路输出错误提示);

　　4.差分约束系统;

　　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。

#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;

int b[210][2],tol;
double hy[210][2],dis[110];

bool Bellman(int s,int n,double V)
{
for(int i=1;i<=n;i++)
dis[i]=0;
dis[s]=V;
for(int i=1;i<n;i++)
{
bool flag=0;
for(int j=0;j<tol;j++)
{
int x1=b[j][0],x2=b[j][1];
if(dis[x2]<(dis[x1]-hy[j][1])*hy[j][0])
{
flag=1;
dis[x2]=(dis[x1]-hy[j][1])*hy[j][0];
}
}
if(flag==0)return 0;
}
for(int i=0;i<tol;i++)
if(dis[b[i][1]]<(dis[b[i][0]]-hy[i][1])*hy[i][0])
return 1;
return 0;
}

int main()
{
double v;
int n,m,s;
while(~scanf("%d%d%d%lf",&n,&m,&s,&v))
{
tol=0;
while(m--)
{
int a1,a2;
double R1,C1,R2,C2;
scanf("%d%d%lf%lf%lf%lf",&a1,&a2,&R1,&C1,&R2,&C2);
b[tol][0]=a1,b[tol][1]=a2;
hy[tol][0]=R1,hy[tol][1]=C1;
tol++;
b[tol][0]=a2,b[tol][1]=a1;
hy[tol][0]=R2,hy[tol][1]=C2;
tol++;
}
if(Bellman(s,n,v))printf("YES\n");
else printf("NO\n");
}
return 0;
}