Currency Exchange
| Time Limit: 1000MS |
|
Memory Limit: 30000K |
| Total Submissions: 11448 |
|
Accepted: 3857 |
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 RAB, CAB, RBA and CBA - 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
以货币为节点，边上有两个值，一个是rate，另一个是commission 显然为有向边 例如<i,j> 表示i to j 的 rate ,commission
题目中最后一段木看明白啥意思
写错了一个字母，改过后才A的
还有就是代码中采用了链式前向星，
前向星是优化spfa的方法之一，可以用数组模拟链表来表示前向星
需要一个索引，至索引头节点，类似于邻接表
/*
链式前向星
int e,d,next[MAXM];
int link[MAXN];
link数组为索引数组，link[i]索引的是i这个节点,可以找到以i为头结点的所有边
e数组建立尾节点表，d数组建立边权表
next数组把所有以i为头结点的边连接起来
用link[i]志向兄弟中的一个，用next数组找到所有的兄弟，用e，d数组读出尾节点和边权
*/
1
#include<stdio.h>
2#include<string.h>
3#include<math.h>
4#define MAX 10000
5int n,num,now;
6double total;
7int s;
8int e[MAX+5],next[MAX+5],link[MAX+5];
9double rate[MAX+5],com[MAX+5];
10short vis[MAX+5];
11double dis[MAX+5];
12int queue[MAX*5],head,tail;
13void add(int a,int b,double k1,double k2)
14
{
15
s++;
16 e[s]=b;
17 rate[s]=k1;
18 com[s]=k2;
19 next[s]=link[a];
20 link[a]=s;
21
}
22void init()
23{
24 int i,a,b;
25 double rab,cab,rba,cba;
26 scanf("%d %d %d %lf",&n,&num,&now,&total);
27 s=0;
28 memset(next,0,sizeof(next));
29 memset(link,0,sizeof(link));
30 for (i=1; i<=num ; i++ )
31
{
32 scanf("%d %d %lf %lf %lf %lf",&a,&b,&rab,&cab,&rba,&cba);
33 add(a,b,rab,cab);
34 add(b,a,rba,cba);
35
}
36}
37short spfa()
38{
39 int i,u,j;
40 memset(vis,0,sizeof(vis));
41 queue[1]=now;
42 for (i=0;i<=n ;i++ )
43 {
44 dis[0]=0;
45 }
46 dis[now]=total;
47 vis[now]=1;
48 head=0;
49 tail=1;
50 while (head<tail)
51 {
52 head++;
53 u=queue[head];
54 j=link[u];
55 while (j!=0)
56 {
57 if ((dis[u]-com[j])*rate[j]>dis[e[j]])
58 {
59 dis[e[j]]=(dis[u]-com[j])*rate[j];
60 if (!vis[e[j]])
61 {
62 vis[e[j]]=1;
63 tail++;
64 queue[tail]=e[j];
65 }
66 }
67 if (dis[now]>total)
68 {
69 return 1;
70 }
71 j=next[j];
72 }
73 vis[u]=0;
74 }
75 return 0;
76}
77int main()
78{
79 init();
80 if (spfa()==1)
81 printf("YES\n");
82 else printf("NO\n");
83 return 0;
84}
85