poj 2240 Arbitrage

又是套汇为题，初始化d数组为1就可以了，用bellman—ford算法。

 1 #include<stdio.h>

 2 #include<string.h>

 3 

 4 int n, m, num, u[2000], v[2000];

 5 double w[2000], d[50], ra;

 6 char s[50][100], s1[100], s2[100];

 7 

 8 void solve()

 9 {

10     for(int i = 0; i <= n; i ++) d[i] = 1;

11     for(int k = 0; k < n-1; k ++)

12         for(int i = 0; i < m; i ++)

13         {

14             int x = u[i], y = v[i];

15             if(d[y] < d[x]*w[i]) d[y] = d[x]*w[i];

16         }    

17     int flag = 0;

18     for(int i = 0; i < m; i++)

19     {

20         int x = u[i],y = v[i];

21         if(d[y] < d[x]*w[i]) flag = 1;

22     }

23     if(flag == 1) printf("Case %d: Yes\n",num++);

24     else printf("Case %d: No\n",num++);

25 }

26 

27 void init()

28 {

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

30     {

31         if(n == 0) break;

32         for(int i = 0; i < n; i ++)

33             scanf("%s",s[i]);

34         scanf("%d",&m);

35         for(int i = 0; i < m; i ++)

36         {

37             scanf("%s",s1);

38             scanf("%lf",&ra);

39             scanf("%s",s2);

40             int a, b;

41             for(int j = 0; j < n; j ++)

42             {

43                 if(!strcmp(s[j], s1)) a = j;

44                 if(!strcmp(s[j], s2)) b = j;

45             }

46             u[i] = a;

47             v[i] = b;

48             w[i] = ra;

49         }

50         solve();

51     }    

52 }

53 

54 int main()

55 {

56     num = 1;

57     init();

58     return 0;

59 }