POJ 2240 Arbitrage (最短路)

Arbitrage

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17145		Accepted: 7238

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. 

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.

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





对应关系用map来处理，然后判断是否存在正环。有一点感悟就是，从A -> B，若存在正环，那么A -> B再B -> A之后A的值就会被变大，再次更新再次变大。

 1 #include <iostream>

 2 #include <string>

 3 #include <map>

 4 #include <vector>

 5 #include <queue>

 6 #include <cstdio>

 7 using    namespace    std;

 8 

 9 const    int    INF = 0xffffff;

10 const    int    SIZE = 35;

11 int    N,M;

12 double    D[SIZE];

13 struct    Node

14 {

15     int    from,to;

16     double    cost;

17 }G[1000];

18 map<string,int>    mapstring;

19 

20 bool    Bellman_Ford(int);

21 bool    relax(int,int,double);

22 int    main(void)

23 {

24     string    box,box_2;

25     double    num;

26     int    from,to;

27     int    count = 0;

28 

29     while(scanf("%d",&N) && N)

30     {

31         count ++;

32         for(int i = 1;i <= N;i ++)

33         {

34             cin >> box;

35             mapstring.insert(pair<string,int>(box,i));

36         }

37         scanf("%d",&M);

38         for(int i = 0;i < M;i ++)

39         {

40             cin >> box >> num >> box_2;

41             G[i].from = mapstring[box];

42             G[i].to = mapstring[box_2];

43             G[i].cost = num;

44         }

45         bool    flag = true;

46         for(int i = 1;i <= N;i ++)

47             if(!Bellman_Ford(i))

48             {

49                 printf("Case %d: Yes\n",count);

50                 flag = false;

51                 break;

52             }

53         if(flag)

54             printf("Case %d: No\n",count);

55     }

56 

57     return    0;

58 }

59 

60 bool    Bellman_Ford(int s)

61 {

62     bool    update;

63     fill(D,D + SIZE,0);

64     D[s] = INF;

65 

66     for(int i = 0;i < N - 1;i ++)

67     {

68         update = false;

69         for(int j = 0;j < M;j ++)

70             if(relax(G[j].from,G[j].to,G[j].cost))

71                 update = true;

72         if(!update)

73             break;

74     }

75     for(int i = 0;i < M;i ++)

76         if(relax(G[i].from,G[i].to,G[i].cost))

77             return    false;

78 

79     return    true;

80 }

81 

82 bool    relax(int from,int to,double cost)

83 {

84     if(D[to] < D[from] * cost)

85     {

86         D[to] = D[from] * cost;

87         return    true;

88     }

89     return    false;

90 }