Currency Exchange

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

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.

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.

If Nick can increase his wealth, output YES, in other case output NO to the output file.

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

YES

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\n");

       else  printf("NO\n");

   }

   return 0;

}

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

#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\n");

         else

         printf("NO\n");

              }

     return 0;

}