【题目描述】

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

【题目分析】
我一开始看到这个题目想的是一个搜索，我想者搜索一遍如果有自环而且回到原点的时候钱变多了就可以。可是看别人博客发现这样是不对的，如果在中间可以通过一个环把钱增多那么就使劲在这个环上把钱变的很多再换成原来的钱肯定是可以的，所以最重要的是判断正环，这里没有用搜索判断是否有正环而是用的Bellman-Ford算法的变形，Bellman-Ford算法可以用来求最短路并判断有无负环，可是如果把这个算法大小符号反过来就可以求有无正环。具体的算法的正确性证明等还不是很了解，有时间应该总结一下，先学会用吧，算法倒是挺简单的。

【AC代码】

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<climits>
#include<queue>
#include<vector>
#include<set>
#include<map>
using namespace std;typedef long long ll;
const int MAXN=105;
struct node
{int s,e; double c,r;
}edge[MAXN<<1],E;
int tot;
int n,m,s;
double v;
double dis[MAXN];bool bellman()
{memset(dis,0,sizeof(dis));dis[s]=v;int T=n-1;bool flag;while(T--){flag=false;for(int i=0;i<tot;i++){E=edge[i];if(dis[E.e]<(dis[E.s]-E.c)*E.r){dis[E.e]=(dis[E.s]-E.c)*E.r;flag=true;}}if(!flag) break;}for(int i=0;i<tot;i++){E=edge[i];if(dis[E.e]<(dis[E.s]-E.c)*E.r){return true;}}return false;
}int main()
{int a,b; double r1,r2,c1,c2;while(~scanf("%d%d%d%lf",&n,&m,&s,&v)){tot=0;for(int i=0;i<m;i++){scanf("%d%d%lf%lf%lf%lf",&a,&b,&r1,&c1,&r2,&c2);edge[tot].s=a; edge[tot].e=b; edge[tot].r=r1; edge[tot].c=c1; tot++;edge[tot].s=b; edge[tot].e=a; edge[tot].r=r2; edge[tot].c=c2; tot++;}if(bellman()){printf("YES\n");}else{printf("NO\n");}}return 0;
}