Arbitrage
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 15640 | Accepted: 6563 |
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.
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.
Input
The input will contain one or more test cases. Om the first line of each test case there is an integer n (1<=n<=30), representing the number of different currencies. The next n lines each contain the name of one currency. Within a name no spaces will appear. The next line contains one integer m, representing the length of the table to follow. The last m lines each contain the name ci of a source currency, a real number rij which represents the exchange rate from ci to cj and a name cj of the destination currency. Exchanges which do not appear in the table are impossible.
Test cases are separated from each other by a blank line. Input is terminated by a value of zero (0) for n.
Test cases are separated from each other by a blank line. Input is terminated by a value of zero (0) for n.
Output
For each test case, print one line telling whether arbitrage is possible or not in the format "Case case: Yes" respectively "Case case: No".
Sample Input
3 USDollar BritishPound FrenchFranc 3 USDollar 0.5 BritishPound BritishPound 10.0 FrenchFranc FrenchFranc 0.21 USDollar 3 USDollar BritishPound FrenchFranc 6 USDollar 0.5 BritishPound USDollar 4.9 FrenchFranc BritishPound 10.0 FrenchFranc BritishPound 1.99 USDollar FrenchFranc 0.09 BritishPound FrenchFranc 0.19 USDollar 0
Sample Output
Case 1: Yes Case 2: No
Source
感觉自己这种算法还是太弱了......唉！ 没有用floy之间用bellman_floy来做的，bellman算法其实总结起来就是三点：
第一： 初始化
第二： 循环优化求解最大或者最小
第三 ： 检测是否存在负环
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<string> 5 #include<iostream> 6 #include<map> 7 #include<iterator> 8 using namespace std; 9 const double inf =-100; 10 const int maxn = 105; 11 struct node 12 { 13 int u,v; 14 double val; 15 }; 16 node edge[1000]; 17 double dist[maxn]; 18 /*松弛状态判别*/ 19 bool relax(int u,int v,double val){ 20 if(dist[v]<dist[u]*val){ 21 dist[v]=dist[u]*val; 22 return 1; 23 } 24 return 0; 25 } 26 bool Bellman(int st,int n,int m){ 27 for(int i=1;i<=n;i++){ //初始化 28 dist[i]=inf; 29 } 30 dist[st]=1; 31 bool flag; 32 /*循环优化部分*/ 33 for(int i=1; i<n;i++) { 34 flag=false; 35 for(int j=1;j<=m;j++){ 36 if(relax(edge[j].u,edge[j].v,edge[j].val)) 37 flag=true; 38 } 39 if(!flag) break; 40 } 41 /*检验部分*/ 42 for(int i=1;i<=m;i++){ 43 if(relax(edge[i].u,edge[i].v,edge[i].val)) 44 return 1; //有负圈 45 } 46 return 0; 47 } 48 49 int main() 50 { 51 int n,m; 52 map<string ,int> sac; 53 string temp; 54 int test=1; 55 while(scanf("%d",&n)==1&&n!=0){ 56 if(!sac.empty())sac.clear(); 57 for(int i=1;i<=n;i++){ 58 cin>>temp; 59 // sac.insert(pair<string ,int>(temp,i)); 60 sac[temp]=i; 61 } 62 cin>>m; 63 double ss; 64 string aa,bb; 65 map<string,int>::iterator p1,p2; 66 for(int i=1 ; i<=m ; i++ ){ 67 cin>>aa>>ss>>bb; 68 p1=sac.find(aa); 69 p2=sac.find(bb); 70 edge[i].u=p1->second; 71 edge[i].v=p2->second; 72 edge[i].val=ss; 73 } 74 75 if(Bellman(1,n,m)) printf("Case %d: Yes ",test); 76 else printf("Case %d: No ",test); 77 test++; 78 // printf("%lf ",dist[1]); 79 } 80 return 0; 81 }