POJ——T1860 Currency Exchange

http://poj.org/problem?id=1860

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 29874		Accepted: 11251

题目大意：

´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。  

 

 题解：

´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。 

所以用SPFA求正环就可以了

 

DFS求环法：

 #include <cstring>
 #include <cstdio>

 #define dou double
 #define INF 1<<29
 #define MAX(a,b) ( a>b ?a :b )

 using namespace std;

 const int N();
 int n,m,s,u,v;
 dou money,uvr,uvl,vur,vul;
 int head[N],sumedge;
 struct Edge
 {
     int to,next;
     dou rate,lose;
     Edge(int to=,int next=,dou rate=0.00,dou lose=0.00) :
         to(to),next(next),rate(rate),lose(lose) {}
 }edge[N<<];

 void ins(int from,int to,dou rate,dou lose)
 {
     edge[++sumedge]=Edge(to,head[from],rate,lose);
     head[from]=sumedge;
 }

 dou change_money(dou x,dou rate,dou lose)
 {     return (x-lose)*rate ;    }

 int vis[N],if_YES;
 dou dis[N];

 void SPFA(int now)
 {
     vis[now]=;
     if(if_YES) return ;
     for(int i=head[now];i;i=edge[i].next)
     {
         int go=edge[i].to;
         dou rate=edge[i].rate,lose=edge[i].lose;
         dou cmoney=change_money(dis[now],rate,lose);
         if(cmoney>dis[go])
         {
             if(vis[go])
             {
                 if_YES=true;
                 break ;
             }
             dis[go]=cmoney;
             SPFA(go);
         }
     }
     vis[now]=;
     return ;
 }

 void init(int n)
 {
     if_YES=sumedge=;
     memset(dis,,sizeof(dis));
     memset(head,,sizeof(head));
 }

 int main()
 {
 //    freopen("made.txt","r",stdin);
 //    freopen("myout.txt","w",stdout);

     while(~scanf("%d%d%d%lf",&n,&m,&s,&money))
     {
         if(s>n)
         {
             printf("NO\n");
             continue;
         }
         init(n);
         for(;m;m--)
         {
             scanf("%d%d%lf%lf%lf%lf",&u,&v,&uvr,&uvl,&vur,&vul);
             ins(u,v,uvr,uvl); ins(v,u,vur,vul);
         }
         dis[s]=money; SPFA(s);
         if(if_YES) printf("YES\n");
         else printf("NO\n");
     }
     return ;
 }

BFS求环法：

 

 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #include<queue>
 using namespace std;
 int u,v,w;
 const int maxn = ;
 const int maxm = ;
 const int oo = <<;
 struct node
 {
     int u;
     int v;
     double x,y ;
     int next;
 }edge[maxm];
 double dis[maxn];
 int m,n,num;
 double ount;
 int head[maxn],cnt,sum[maxn];
 int vis[maxn] = {};
 queue<int>qu;
 void add(int u,int v,double x,double y)
 {
     edge[cnt].u = u ;
     edge[cnt].v = v ;
     edge[cnt].x = x ;
     edge[cnt].y = y ;
     edge[cnt].next = head[u];
     head[u] = cnt++ ;
 }
 int spfa(int s)
 {
     for(int i =  ; i < m ; i++)
     {
         dis[i] = ;
         vis[i] =  ;
     }
     dis[s] = ount;
     qu.push(s);
     vis[s] =  ;
     while(!qu.empty())
     {
         int u = qu.front();
         qu.pop();
         vis[u] = ;
         for(int i = head[u] ; i != - ; i = edge[i].next)
         {
             int v = edge[i].v;
             if((dis[u]-edge[i].y)*edge[i].x > dis[v])
             {
                 dis[v] = (dis[u]-edge[i].y)*edge[i].x;
                 if(!vis[v])
                 {
                     vis[v] = ;
                     qu.push(v);
                 }
                 sum[v]++;
                 if(sum[v] > m)
                 return -;
             }
         }
     }
     return ;
 }
 void Init()
 {
     cnt =  ;
     memset(head,-,sizeof(head));
     memset(sum,,sizeof(sum));
     while(!qu.empty())
     qu.pop();
 }
 int main()
 {
     freopen("made.txt","r",stdin);
     freopen("stdout.txt","w",stdout);

     while(scanf("%d %d %d %lf",&m,&n,&num,&ount)!=EOF)
     {
         Init();
         int u,v;
         double x,y,xx,yy ;
         for(int i =  ; i < n ; i++)
         {
             scanf("%d %d %lf %lf %lf %lf",&u,&v,&x,&y,&xx,&yy);
             add(u,v,x,y);
             add(v,u,xx,yy);
         }
         if(spfa(num) > )
         cout<<"NO"<<endl;
         else cout<<"YES"<<endl;
     }
     return ;
 }