poj1860 bellman—ford队列优化 Currency Exchange

Currency Exchange

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 22123		Accepted: 7990

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

Source

 

 

 

 

解析

题意：

         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

怎么找正权回路呢？（正权回路：在这一回路上，顶点的权值能不断增加即能一直进行松弛）

分析：

一种货币就是一个点

一个“兑换点”就是图上两种货币之间的一个兑换方式，是双边，但A到B的汇率和手续费可能与B到A的汇率和手续费不同。

唯一值得注意的是权值，当拥有货币A的数量为V时，A到A的权值为K，即没有兑换

而A到B的权值为(V-Cab)*Rab

本题是“求最大路径”，之所以被归类为“求最小路径”是因为本题题恰恰与bellman-Ford算法的松弛条件相反，求的是能无限松弛的最大正权路径，但是依然能够利用bellman-Ford的思想去解题。

因此初始化dis(S)=V   而源点到其他点的距离（权值）初始化为无穷小（0），当s到其他某点的距离能不断变大时，说明存在最大路径；如果可以一直变大，说明存在正环。判断是否存在环路，用Bellman-Ford和spfa都可以。

spfa算法：

下面是bellman——ford队列优化的代码

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<queue>
#include<algorithm>
using namespace std;
double cost[105][105],rate[105][105];
int n,vis[105];
double v,dis[105];
bool bellman_ford(int start){
   memset(dis,0,sizeof(dis));
   memset(vis,0,sizeof(vis));
   dis[start]=v;
   queue<int>q;
   q.push(start);
   vis[start]=1;
   while(!q.empty()){
      int x=q.front();
      q.pop();
      vis[x]=0;
      for(int i=1;i<=n;i++){
         if(dis[i]<(dis[x]-cost[x][i])*rate[x][i]){
             dis[i]=(dis[x]-cost[x][i])*rate[x][i];
             if(dis[start]>v)
             return true;
             if(!vis[i]){
             q.push(i);
               vis[i]=1;
             }
         }
      }
   }
   return false;
}
int main(){
   int m,s;
   while(scanf("%d%d%d%lf",&n,&m,&s,&v)!=EOF){
       memset(cost,0,sizeof(vis));
       memset(rate,0,sizeof(rate));

       for(int i=1;i<=n;i++){
          for(int j=1;j<=n;j++)
          if(i==j)
          rate[i][j]=1.0;
       }
       int x,y;
       double rab,rba,cab,cba;
       for(int i=1;i<=m;i++){
          cin>>x>>y>>rab>>cab>>rba>>cba;
          cost[x][y]=cab;
          cost[y][x]=cba;
          rate[x][y]=rab;
          rate[y][x]=rba;
       }
       if(bellman_ford(s))
       printf("YES
");
       else  printf("NO
");
   }
   return 0;
}

　　下面是bellman——ford算法

bellman——ford算法中的调用函数的解析

如果上一步循环中中途退出，说明不在进行松弛了，那么这一步也不会再次进行松弛
   //上一步不再进行松弛其实是说明不在会有正权环了，如果仍然有正权环还会继续进行松弛，
   //没有正权环其实本题也是输出NO了，如果有正权环，说明可以不断循环增加自己本身的财产，
   //那么及时多循环多少次仍然可以增加自己的收入

代码

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int Count,n,m,s;
double v;
double dis[105];
struct node{
  int x;
  int y;
  double cost,rate;
}que[105];
bool Bellman_Ford(){
   memset(dis,0,sizeof(dis));//此处与Bellman-Ford的处理相反，初始化为源点到各点距离0，到自身的值为原值
   dis[s]=v;
   int flag;
   for(int i=1;i<n;i++){
       flag=0;
      for(int j=0;j<Count;j++){
         int x=que[j].x;
         int y=que[j].y;
         double cost=que[j].cost;
         double rate=que[j].rate;
         if(dis[y]<(dis[x]-cost)*rate){
              dis[y]=(dis[x]-cost)*rate;
              flag=1;
         }
      }
      if(!flag)
      break;
   }
   for(int i=0;i<Count;i++){//正环能够无限松弛,
         if(dis[que[i].y]<(dis[que[i].x]-que[i].cost)*que[i].rate)
         return true;
   }//如果上一步循环中中途退出，说明不在进行松弛了，那么这一步也不会再次进行松弛
   //上一步不再进行松弛其实是说明不在会有正权环了，如果仍然有正权环还会继续进行松弛，
   //没有正权环其实本题也是输出NO了，如果有正权环，说明可以不断循环增加自己本身的财产，
   //那么及时多循环多少次仍然可以增加自己的收入

   return false;
}
int main(){
     while(scanf("%d%d%d%lf",&n,&m,&s,&v)!=EOF){
         int x,y;
         double rab,rba,cba,cab;
         Count=0;
         for(int i=1;i<=m;i++){
             scanf("%d%d%lf%lf%lf%lf",&x,&y,&rab,&cab,&rba,&cba);
             que[Count].x=x;
             que[Count].y=y;
             que[Count].cost=cab;
             que[Count].rate=rab;
             Count++;
             que[Count].x=y;
             que[Count].y=x;
             que[Count].cost=cba;
             que[Count].rate=rba;
             Count++;
         }
         if(Bellman_Ford())
         printf("YES
");
         else
         printf("NO
");
              }
     return 0;
}