用Bellman-Ford最短路径算法，判断是否存在正环，如果存在，说明可以无限增加金钱（你不停绕着正环走就能无限增加）。如果不存在，就输出NO；
Bellman-Ford是求最短路并判断是否有负环，也能求是否有正环。因为有n个点的图，最优路径最多n-1个路径，所以从起点出发松弛n-1次就行。如果第n
次还有路径变长，就说明存在正环。我们直接松弛n次就好（就是把每条路径判断n次，看是否能优化），如果发现没有路径可优化，说明不存在正环，返回
0就行。
#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iomanip>
#define ll long long
#define ld long double
#define INF 0x3f3f3f3f
using namespace std;
struct node
{
int x,y;
double cost,v;
} G[202]; //用来储存边，包括端点，转换率和消耗
int n,m,X,edgenum=0; //边的条数
double V;
double dis[102];
bool ford()
{
memset(dis,0,sizeof dis);
dis[X]=V;
for(int j=0; j<n; j++) //直接进行n次松弛
{
int flag=0; //判断有没有路径被优化
for(int i=0; i<edgenum; i++)
{
if((dis[G[i].x]-G[i].cost)*G[i].v>dis[G[i].y])
{
dis[G[i].y]=(dis[G[i].x]-G[i].cost)*G[i].v;
flag=1;
}
}
if(flag==0) //没有路径被优化，说明不存在正环
return 0;
}
return 1;
}
int main()
{
//std::ios::sync_with_stdio(false);
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
cin>>n>>m>>X>>V;
int a,b;
double c,d,e,f;
for(int i=0; i<m; i++)
{
cin>>a>>b>>c>>d>>e>>f;
G[edgenum].x=a;
G[edgenum].y=b;
G[edgenum].v=c;
G[edgenum].cost=d;
edgenum++;
G[edgenum].x=b;
G[edgenum].y=a;
G[edgenum].v=e;
G[edgenum].cost=f;
edgenum++;
}
if(ford())
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
return 0;
}
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
