exchange服务器之Currency Exchange-职坐标

exchange服务器之Currency Exchange

白羽 2018-12-04 来源：网络阅读 561 评论 0

摘要：本文将带你了解exchange服务器之Currency Exchange，希望本文对大家学Exchange有所帮助。

本文将带你了解exchange服务器之Currency Exchange，希望本文对大家学Exchange有所帮助。

Currency Exchange

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 60000/30000K (Java/Other)

Total Submission(s) : 76 Accepted Submission(s) : 21

Problem 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

 这题是说给n种钱，m个换钱点，每个换钱点可以把你手中的前扣掉手续费之后乘以汇率换成另一种，钱b =（钱a - 手续费c）*汇率r 。问给你开始给你v块种类为s的钱，有没有可能通过换钱还钱之后把钱变多。
 其实这题很简单的思想就是进行广搜，用手中的钱通过每个换钱点兑换，广搜下去直到某个时候种类为s且价值大于v就“YES”了，到不了就“NO”，但是广搜超时。把思路回来，每个换钱点都是一条路，用Ford算法求单源最短路径的办法，判断是否有正环，即dis【s】>v的情况。当然，SPFA也一样。至于哪个好，我都写了但我不知道。
代码如下：
//Ford AC#include<iostream>#include<cstring>using namespace std;struct Edge{ int x,y; double rate,cost;}edge[220];int n,m,S,flag;double V;double dis[110];int ioan(){ memset(dis,0,sizeof(dis)); dis[S]=V; int x,y; double r,c; for(int i=1;i<=n-1;i++) { int f=0; for(int j=1;j<=2*m;j++) { x=edge[j].x; y=edge[j].y; r=edge[j].rate; c=edge[j].cost; if(dis[y]<(dis[x]-c)*r) { dis[y]=(dis[x]-c)*r; f=1; } } if(f==0) { return 0; } } for(int j=1;j<=2*m;j++) { x=edge[j].x; y=edge[j].y; r=edge[j].rate; c=edge[j].cost; if(dis[y]<(dis[x]-c)*r) { flag=1; return 1; } } return 0;}int main(){ int x,y; double ra,rb,ca,cb; while(cin>>n>>m>>S>>V) { int k=1; flag=0; for(int i=1;i<=m;i++) { cin>>x>>y>>ra>>ca>>rb>>cb; edge[k].x=x; edge[k].y=y; edge[k].rate=ra; edge[k].cost=ca; k++; edge[k].x=y; edge[k].y=x; edge[k].rate=rb; edge[k].cost=cb; k++; } int f=ioan(); if(f==1) cout<<"YES"<<endl; else cout<<"NO"<<endl; } return 0;} //SPFA AC#include<iostream>#include<cstring>#include<queue>using namespace std;int n,m,s,x,y,flag;double v,ra,rb,ca,cb;double rate[510][510];double cost[510][510];int vis[510];double dis[510];int ioan(){ queue<int>q; memset(vis,0,sizeof(vis)); memset(dis,0,sizeof(dis)); q.push(s); dis[s]=v; vis[s]=1; while(!q.empty()) { int a=q.front(); q.pop(); vis[a]=0; for(int i=1;i<=n;i++) { if(dis[i]<(dis[a]-cost[a][i])*rate[a][i]) { dis[i]=(dis[a]-cost[a][i])*rate[a][i]; if(dis[s]>v)//结束条件，这是当出现先了符合题意的点时直接的结果，，如果到不了的话，知道循环结束拉倒 { flag=1; return 1; } if(vis[i]==0) { vis[i]=1; q.push(i); } } } } return 0;}int main(){ while(cin>>n>>m>>s>>v) { flag=0; memset(rate,0,sizeof(rate)); memset(cost,0,sizeof(cost)); for(int i=1;i<=n;i++) rate[i][i]=1; for(int i=1;i<=m;i++) { cin>>x>>y; cin>>ra>>ca>>rb>>cb; rate[x][y]=ra; rate[y][x]=rb; cost[x][y]=ca; cost[y][x]=cb; } int f=ioan(); if(f==1) cout<<"YES"<<endl; else cout<<"NO"<<endl; } return 0;}