Simple POJ 2240 ARBITRAGE,SPFA.

Description

Arbitrage is the use of discrepancies in currency exchange rates to transform one unit of a currency into more than one UN It's the same currency. For example, suppose that 1 US Dollar buys 0.5 British pound, 1 British pound buys 10.0 French francs, and 1 French franc Buys 0.21 US dollar. Then, by converting currencies, a clever trader can start with 1 US dollar and buy 0.5 * 10.0 * 0.21 = 1.05 US dollars, MA King a profit of 5 percent.

Your job is to write a program this takes a list of currency exchange rates as input and then determines whether arbitrage is possible or not.

The question is whether the currency can be converted and let itself increase ...

Use the SPFA to judge the ring ... Enumerate each point for SPFA (i.e. Floyd is also possible ...). )。。。

The code is as follows:

#include <iostream>#include<cstring>#include<cstdio>#include<queue>using namespacestd;Const intinf=10e8;Const intmaxn= +;structedge{intv; DoubleCost ; Edge (int_v=0,Double_cost=0): V (_v), cost (_cost) {}};vector<Edge>E[MAXN];BOOLVIS[MAXN];intCOUNODE[MAXN];BOOLSPFA (DoubleLowcost[],intNintstart) {Queue<int>que; intu,v; DoubleC; intLen;  for(intI=1; i<=n;++i) {vis[i]=0; Counode[i]=0; Lowcost[i]=0; } Vis[start]=1; Counode[start]=1; Lowcost[start]=1;    Que.push (start);  while(!Que.empty ()) {u=Que.front ();        Que.pop (); Vis[u]=0; Len=e[u].size ();  for(intI=0; i<len;++i) {v=e[u][i].v; C=E[u][i].cost; if(lowcost[u]*c>Lowcost[v]) {Lowcost[v]=lowcost[u]*C; if(!Vis[v]) {Vis[v]=1; ++Counode[v];                    Que.push (v); if(counode[v]>=N)return 0; }            }        }    }    return 1;} InlinevoidAddedge (intUintVDoublec) {E[u].push_back (Edge (V,c));}Charss[ +][ -];DoubleANS[MAXN];intN;intFindChar*R) {     for(intI=1; i<=n;++i)if(strcmp (s,ss[i]) = =0)            returni;}intMain () {intM; BOOLOK; Charts1[ -],ts2[ -]; intt1,t2; DoubleTR; intcas=1;  for(SCANF ("%d", &n); N;SCANF ("%d", &n), + +CAs) {         for(intI=1; i<=n;++i) {scanf ("%s", Ss[i]);        E[i].clear (); } scanf ("%d",&l);  for(intI=1; i<=m;++i) {scanf ("%s%lf%s",ts1,&tr,ts2); T1=find (TS1); T2=find (TS2);        Addedge (T1,T2,TR); } OK=0;  for(intI=1; i<=n;++i)if(!SPFA (ans,n,i)) {OK=1;  Break; } printf ("Case %d:", CAs); if(OK) printf ("yes\n"); Elseprintf ("no\n"); }    return 0;}

View Code

Simple POJ 2240 ARBITRAGE,SPFA.