[2016-04-13] [POJ] [1860] [Currency Exchange]

• Time: 2016-04-13-23:48:46
• Topic number: [2016-04-13][poj][1860][currency Exchange]
• The main idea: Currency exchange, ask the last can be changed by the way make money more,
• Analysis:
  • Direct SPFA Determine if there is a ring, and if so, it can add value indefinitely.
  • If there is no positive ring, then directly determine whether the final d[s] is greater than the initial value

  
 

   
  

  
#include<cstdio>
   
  

  
#include<vector>
   
  

  
#include<cstring>
   
  

  
#include<queue>
   
  

  
using namespace std;
   
  

  
const int maxn = 100 + 10;
   
  

  
struct Edge{
   
  

  
  int   ;  double   c    R    
   
  

  
  edge   ( int   _v  =    0   double   _c  =    0  ,  Span class= "KWD" >double   _r  =    Span class= "lit" >0 ):  v   ( _v  ),  c   ( _c   r   ( _r  Span class= "pun") {}   
   
  

  
};
   
  

  
vector<Edge> e[maxn];
   
  

  
void ini(int n ){
   
  

  
 for(int i = 0 ; i <= n ; ++i){
   
  

  
 e[i].clear();
   
  

  
 }
   
  

  
}
   
  

  
void addedge(int u,int v,double c,double r){
   
  

  
 e[u].push_back(Edge(v,c,r));
   
  

  
}
   
  

  
int cnt[maxn];double d[maxn];
   
  

  
bool spfa(int s,int n,double v){
   
  

  
  for   ( int   i  =    0    ;   I  <=   n     ++  i  )   D  [ i      =    0    
   
  

  
 d[s] = v;
   
  

  
 queue<int> q;q.push(s);
   
  

  
 memset(cnt, 0 , sizeof(cnt));
   
  

  
 cnt[s] = 1;
   
  

  
 while(!q.empty()){
   
  

  
 int u = q.front();
   
  

  
 q.pop();
   
  

  
  for   ( int   i  =    0    ;   I  <   e  [ u  .  size   i    
   
  

  
  int   v  =   e  [ u  ][ i  .  v    
   
  

  
  if   ( d  [ v      <     ( d  [ u      -  e  [ u  ][ i   c     *   e  [ u  ][ i   r     
   
  

  
 D  [v  ]    =  Span class= "PLN" >   ( d  [ u      -  e  [ u  ][ i   c     *   e  [ u  ][ i   r    
   
  

  
 q.push(v);
   
  

  
 if(++cnt[v] > n) return true;
   
  

  
 }
   
  

  
 }
   
  

  
 }
   
  

  
 return d[s] > v;
   
  

  
}
   
  

  
int main(){
   
  

  
 int n,m,s;
   
  

  
 double v;
   
  

  
 scanf("%d%d%d%lf",&n,&m,&s,&v);
   
  

  
  for   ( int   i  =    0    ;   I  <   m     ++  i  ) {  
   
  

  
  int    b   double    rab   cab  Span class= "pun",  rba   CBA  Span class= "pun";   
   
  

  
 scanf("%d%d%lf%lf%lf%lf",&a,&b,&Rab,&Cab,&Rba,&Cba);
   
  

  
 addedge(a,b,Cab,Rab);
   
  

  
 addedge(b,a,Cba,Rba);
   
  

  
 }
   
  

  
 puts(spfa(s,n,v)?"YES":"NO");
   
  

  
 return 0;
   
  

  
}
  
 

From for notes (Wiz)

[2016-04-13] [POJ] [1860] [Currency Exchange]