【題目描述】

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

【題目分析】
我一開始看到這個題目想的是一個搜索，我想者搜索一遍如果有自環而且回到原點的時候錢變多了就可以。可是看別人博客發現這樣是不對的，如果在中間可以通過一個環把錢增多那么就使勁在這個環上把錢變的很多再換成原來的錢肯定是可以的，所以最重要的是判斷正環，這里沒有用搜索判斷是否有正環而是用的Bellman-Ford算法的變形，Bellman-Ford算法可以用來求最短路并判斷有無負環，可是如果把這個算法大小符號反過來就可以求有無正環。具體的算法的正確性證明等還不是很了解，有時間應該總結一下，先學會用吧，算法倒是挺簡單的。

【AC代碼】

#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;
}