poj1860--Currency Exchange

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Bellman-ford算法的反向应用－－正循环检查

/** \brief poj 1860 Bellman-Ford * * \param date 2014/7/24 * \param state AC * \return memory 708K time 141ms * */#include <iostream>#include <fstream>#include <cstring>using namespace std;struct RateAndCom{//public:    int a;    int b;    double rate;    double Com;};//Map[MAXN];const int MAXN=101;RateAndCom Map[101*2];double dis[MAXN];int N;//货币种数int M;//兑换点数量int S;//持有第s种货币double V;//第s种货币本金int allEdge;bool Bellman_Ford(){    memset(dis,0,sizeof(dis));    dis[S]=V;    /*relax*/    bool flag;    for(int i=1;i<=N-1;i++)    {        flag=false;        for(int j=0;j<allEdge;j++)            if(dis[Map[j].b] < (dis[Map[j].a]-Map[j].Com)*Map[j].rate)        {            dis[Map[j].b] = (dis[Map[j].a]-Map[j].Com)*Map[j].rate;            flag=true;        }        if(!flag)            break;    }    for(int k=0;k<allEdge;k++)        if(dis[Map[k].b] < (dis[Map[k].a]-Map[k].Com)*Map[k].rate)        return true;    return false;}int main(){    //cout << "Hello world!" << endl;    //freopen("input.txt","r",stdin);    //while(scanf("%d %d %d %f",&N,&M,&S,&V)!=EOF)    while(cin>>N>>M>>S>>V)    {        allEdge=0;        for(int i=0;i<M;i++)        {            int a,b;            double Rab;            double Cab;            double Rba;            double Cba;            //cin>>a>>b>>Map[a][b].rate>>Map[a][b].Commission            //>>Map[b][a].rate>>Map[b][a].Commission;            cin>>a>>b>>Rab>>Cab>>Rba>>Cba;            Map[allEdge].a=a;            Map[allEdge].b=b;            Map[allEdge].rate=Rab;            Map[allEdge].Com=Cab;            allEdge++;            Map[allEdge].a=b;            Map[allEdge].b=a;            Map[allEdge].rate=Rba;            Map[allEdge].Com=Cba;            allEdge++;        }        //Bellman-Ford        if(Bellman_Ford())            cout<<"YES"<<endl;        else cout<<"NO"<<endl;    }    return 0;}

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