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\n"); else printf("No\n"); } return 0; }
时间: 2024-10-12 09:20:57