POJ 1860 bellman_ford算法

最新推荐文章于 2021-10-18 20:35:06 发布

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#算法 #exchange #numbers #each #integer #output

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本文介绍了一道关于货币兑换的最短路径问题，通过构建复杂的图结构并利用Bellman-Ford算法来解决。问题旨在判断是否能通过一系列兑换操作增加初始资本。

一道最短路的题，建图有点难度，想了一个多小时的建图，然后再用bellman_ford就可以轻松水过了。。。。。。。。题目：

Currency Exchange

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 11990		Accepted: 4040

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 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<=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

ac代码：

#include <iostream>
#include <cstdio>
#include <string.h>
using namespace std;
int numcoin,numstation,start;
const int N=110;
#define MIN -0x7fffffff
double value;
double rate[N][N],fee[N][N];
int visted[N];
double dis[N];
struct edge{
  int leftpoint,rightpoint;
  double rate,fee;
}ee[2*N];
bool bellman_ford(){
  memset(visted,0,sizeof(visted));
  for(int i=0;i<N;++i)
	  dis[i]=MIN;
  int pos=start;
  dis[pos]=value;
  bool flag=0;
  for(int i=1;i<=numcoin;++i){
    flag=0;
	for(int j=0;j<=2*numstation;++j){
		if(dis[ee[j].leftpoint]!=MIN&&dis[ee[j].rightpoint]<(dis[ee[j].leftpoint]-ee[j].fee)*ee[j].rate){
		  flag=1;
		  dis[ee[j].rightpoint]=(dis[ee[j].leftpoint]-ee[j].fee)*ee[j].rate;
		}
	}
	if(!flag)
		break;
  }
  if(!flag)
	  return false;
  return true;
}
int main(){
	//freopen("1.txt","r",stdin);
	while(~scanf("%d%d%d%lf",&numcoin,&numstation,&start,&value)){
	  //totalvalue=eachvalue*numcount;
	  int x,y;
	  double rate1,fee1,rate2,fee2;
	  for(int i=0;i<numstation;++i){
	    scanf("%d%d%lf%lf",&x,&y,&rate1,&fee1);
		ee[i].leftpoint=x;ee[i].rightpoint=y;
		ee[i].rate=rate1;ee[i].fee=fee1;
		scanf("%lf%lf",&rate2,&fee2);
		ee[i+numstation].leftpoint=y;ee[i+numstation].rightpoint=x;
		ee[i+numstation].rate=rate2;ee[i+numstation].fee=fee2;
	  }
	 bool flag=bellman_ford();
	 if(flag)
		 printf("YES\n");
	 else
		 printf("NO\n");
	}
  return 0;
}