HDU 1860 - Currency Exchange(最短路)
来源:互联网 发布:nsp网络 编辑:程序博客网 时间:2024/06/10 02:17
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.
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
题意:有多种汇币,汇币之间可以相互转换,公式为 a = (b - c) * r ,求出是否存在正权回路,且货币数增加。
思路:使用ford 的判圈方法。总共需要循环所有的边n - 1次,若第n - 1次时仍然会存在d【i】增加,则说明存在圈。
CODE:
#include <iostream>#include <cstdio>#include <queue>#include <algorithm>using namespace std;const int MAX = 1005;struct edge{ int to; double r, c; };vector<edge> g[MAX];int n, m, st;double v, d[1005];void ford(int s){ for(int i = 0; i < n; ++i) { d[i] = 0.0; } d[s] = v; for(int i = 0; i < n; ++i) { for(int ii = 1; ii <= n; ++ii) for(int j = 0; j < g[ii].size(); ++j) { edge e = g[ii][j]; if((d[ii] - e.c) * e.r > d[e.to]) { d[e.to] = (d[ii] - e.c) * e.r; if(i == n - 1) { printf("YES\n"); return; } } } } printf("NO\n"); return;}int main(){//freopen("in", "r", stdin); while(~scanf("%d %d %d %lf", &n, &m, &st, &v)) { for(int i = 0; i < n; ++i) { g[i].clear(); } int a, b; double r1, c1, r2, c2; edge e; while(m--) { scanf("%d %d %lf %lf %lf %lf", &a, &b, &r1, &c1, &r2, &c2); e.to = b; e.r = r1; e.c = c1; g[a].push_back(e); e.to = a; e.r = r2; e.c = c2; g[b].push_back(e); } ford(st); } return 0;}
0 0
- HDU 1860 - Currency Exchange(最短路)
- hdu 1860 Currency Exchange(最短路、深搜spfa)
- poj 1860 -- Currency Exchange(最短路)
- poj 1860 Currency Exchange (最短路)
- POJ 1860 Currency Exchange(最短路)
- POJ 1860 Currency Exchange(最短路SPFA)
- POJ 1860Currency Exchange BFS最短路
- PKU 1860 Currency Exchange 最短路 bellman
- POJ - 1860 Currency Exchange(最短路)
- poj 1860 Currency Exchange(最短路)
- poj 1860Currency Exchange(bellman 最短路)
- POJ 1860 Currency Exchange(最短路)
- POJ-1860-Currency Exchange [最短路][BellmanFord]
- 最短路(Bellman_ford) Currency Exchange
- Currency Exchange(最短路_Beelman_Ford)
- Poj_1860 Currency Exchange(最短路)
- POJ 1860 Currency Exchange(最短路 Bellman-Ford)
- poj 1860 Currency Exchange (最短路变形-spfa)
- WPF之DataGrid应用
- android 布局相关知识
- eclipse 解决乱码问题
- 第十七周项目7 电子词典(还有疑问,求解答)
- SurfaceView实现视频播放
- HDU 1860 - Currency Exchange(最短路)
- soaplib之model?
- ssh连接linux
- IO的阻塞与非阻塞、同步与异步以及Java网络IO交互方式
- Change checkbox size WPF
- linux的NFS详细配置方法
- 程序员如何快速准备面试中的算法
- awk 用法(全)
- 工欲善其事,必先利其器-ecplise配置和优化