poj1860 bellman—ford隊列優化 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<=10 3.
For each point exchange rates and commissions are real, given with at most two digits after the decimal point, 10 -2<=rate<=10 2, 0<=commission<=10 2.
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 10 4.

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

題意:

         有多種匯幣,匯幣之間可以交換,這需要手續費,當你用100A幣交換B幣時,A到B的匯率是29.75,手續費是0.39,那么你可以得到(100 - 0.39) * 29.75 = 2963.3975 B幣。問s幣的金額經過交換最終得到的s幣金額數能否增加

貨幣的交換是可以重復多次的,所以我們需要找出是否存在正權回路,且最后得到的s金額是增加的

怎么找正權回路呢?(正權回路:在這一回路上,頂點的權值能不斷增加即能一直進行松弛)


分析:

一種貨幣就是一個點

一個“兌換點”就是圖上兩種貨幣之間的一個兌換方式,是雙邊,但A到B的匯率和手續費可能與B到A的匯率和手續費不同。

唯一值得注意的是權值,當擁有貨幣A的數量為V時,A到A的權值為K,即沒有兌換

而A到B的權值為(V-Cab)*Rab

本題是“求最大路徑”,之所以被歸類為“求最小路徑”是因為本題題恰恰與bellman-Ford算法的松弛條件相反,求的是能無限松弛的最大正權路徑,但是依然能夠利用bellman-Ford的思想去解題。

因此初始化dis(S)=V   而源點到其他點的距離(權值)初始化為無窮小(0),當s到其他某點的距離能不斷變大時,說明存在最大路徑;如果可以一直變大,說明存在正環。判斷是否存在環路,用Bellman-Ford和spfa都可以。

spfa算法:

下面是bellman——ford隊列優化的代碼

#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\n");
       else  printf("NO\n");
   }
   return 0;
}

 

  下面是bellman——ford算法

bellman——ford算法中的調用函數的解析

如果上一步循環中中途退出,說明不在進行松弛了,那么這一步也不會再次進行松弛
   //上一步不再進行松弛其實是說明不在會有正權環了,如果仍然有正權環還會繼續進行松弛,
   //沒有正權環其實本題也是輸出NO了,如果有正權環,說明可以不斷循環增加自己本身的財產,
   //那么及時多循環多少次仍然可以增加自己的收入

 

代碼

 

#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));//此處與Bellman-Ford的處理相反,初始化為源點到各點距離0,到自身的值為原值
   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++){//正環能夠無限松弛,
         if(dis[que[i].y]<(dis[que[i].x]-que[i].cost)*que[i].rate)
         return true;
   }//如果上一步循環中中途退出,說明不在進行松弛了,那么這一步也不會再次進行松弛
   //上一步不再進行松弛其實是說明不在會有正權環了,如果仍然有正權環還會繼續進行松弛,
   //沒有正權環其實本題也是輸出NO了,如果有正權環,說明可以不斷循環增加自己本身的財產,
   //那么及時多循環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\n");
         else
         printf("NO\n");
              }
     return 0;
}

  


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