(简单) 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 unit of 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, making a profit of 5 percent.

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

　　题目就是问货币能不能通过转换而让自己增加。。。

　　用的SPFA来判断的环。。。枚举每一个点进行SPFA (也就是说floyd也是可以的。。。)。。。

代码如下：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>

using namespace std;

const int INF=10e8;
const int MaxN=40;

struct Edge
{
    int v;
    double cost;

    Edge(int _v=0,double _cost=0):v(_v),cost(_cost) {}
};

vector <Edge> E[MaxN];
bool vis[MaxN];
int couNode[MaxN];

bool SPFA(double lowcost[],int n,int start)
{
    queue <int> que;
    int u,v;
    double c;
    int len;

    for(int i=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(int i=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;
}

inline void addEdge(int u,int v,double c)
{
    E[u].push_back(Edge(v,c));
}

char ss[40][100];
double ans[MaxN];
int N;

int find(char *s)
{
    for(int i=1;i<=N;++i)
        if(strcmp(s,ss[i])==0)
            return i;
}

int main()
{
    int M;
    bool ok;
    char ts1[100],ts2[100];
    int t1,t2;
    double tr;
    int cas=1;

    for(scanf("%d",&N);N;scanf("%d",&N),++cas)
    {
        for(int i=1;i<=N;++i)
        {
            scanf("%s",ss[i]);
        
            E[i].clear();
        }

        scanf("%d",&M);

        for(int i=1;i<=M;++i)
        {
            scanf("%s %lf %s",ts1,&tr,ts2);
            t1=find(ts1);
            t2=find(ts2);

            addEdge(t1,t2,tr);
        }

        ok=0;

        for(int i=1;i<=N;++i)
            if(!SPFA(ans,N,i))
            {
                ok=1;
                break;
            }

        printf("Case %d: ",cas);

        if(ok)
            printf("Yes
");
        else
            printf("No
");

    }

    return 0;
}