POJ 1860 Currency Exchange
来源:互联网 发布:淘宝搜索广告位价格 编辑:程序博客网 时间:2024/06/03 18:48
DescriptionView Code
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.
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.
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 20.01 2 1.00 1.00 1.00 1.002 3 1.10 1.00 1.10 1.00
Sample Output
YES
思路:
有多种汇币,汇币之间可以交换,这需要手续费,当你用100A币交换B币时,A到B的汇率是29.75,手续费是0.39,那么你可以得到(100 - 0.39) * 29.75 = 2963.3975 B币。问s币的金额经过交换最终得到的s币金额数能否增加。 货币的交换是可以重复多次的,所以我们需要找出是否存在正权回路,且最后得到的s金额是增加的。
#include<queue>#include<cstdio>#include<cstring>#include<iostream>#define MAXN 110using namespace std;double f[MAXN][MAXN],h[MAXN][MAXN],v,dis[MAXN];int sum[MAXN],n,m,s;bool b[MAXN];inline void read(int&x) { x=0;int f=1;char c=getchar(); while(c>'9'||c<'0') {if(c=='-') f=-1;c=getchar();} while(c>='0'&&c<='9') {x=(x<<1)+(x<<3)+c-48;c=getchar();} x=x*f;}queue<int> q;inline bool spfa(int x) { memset(b,false,sizeof b); memset(dis,0,sizeof dis); memset(sum,0,sizeof sum); q.push(x); dis[x]=v; b[x]=true; while(!q.empty()) { int u=q.front(); q.pop(); b[u]=false; if(dis[x]>v) return true; for(int i=1;i<=n;i++) { if((dis[u]-h[u][i])*f[u][i]>dis[i]) { dis[i]=(dis[u]-h[u][i])*f[u][i]; if(!b[i]) { q.push(i); sum[i]++; b[i]=true; if(sum[i]>n) return true; } } } } return false;}int main() { while(scanf("%d%d%d%lf",&n,&m,&s,&v)!=EOF) { int x,y; for(int i=1;i<=m;i++) { read(x);read(y); scanf("%lf%lf%lf%lf",&f[x][y],&h[x][y],&f[y][x],&h[y][x]); } if(spfa(s)) printf("YES\n"); else printf("NO\n"); while(!q.empty()) q.pop(); } return 0;}
阅读全文
0 0
- POJ 1860 Currency Exchange
- poj 1860 Currency Exchange
- POJ 1860 Currency Exchange
- Poj 1860 Currency Exchange
- POJ 1860 Currency Exchange
- poj 1860 Currency Exchange
- POJ 1860 Currency Exchange
- poj 1860 Currency Exchange
- POJ 1860 Currency Exchange
- POJ 1860 Currency Exchange
- POJ 1860 Currency Exchange
- poj 1860 Currency Exchange
- POJ 1860 Currency Exchange
- poj 1860 Currency Exchange
- POJ-1860-Currency Exchange
- POJ 1860 Currency Exchange
- poj 1860 Currency Exchange
- POJ 1860 Currency Exchange
- 无聊写的压位高精加
- 2030 行
- POJ 2421 Constructing Roads
- 输入三角形的三边长输出面积
- POJ 1679 The Unique MST(次短生成树)
- POJ 1860 Currency Exchange
- POJ 1797 Heavy Transportation
- POJ 2186 Popular Cows
- 1961 Period
- cordova 图标设置
- POJ 2553 The Bottom of a Graph
- 洛谷 P1548 棋盘问题
- Python的hasattr() getattr() setattr() 函数使用方法详解
- 洛谷 P1313 计算系数