Arbitrage
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 9210 |
|
Accepted: 3920 |
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
总结：跟1860相似，但是这个没有确定的值，不明白为什么按照1860的代码会wa，找了题解后明白是求是否存在负环回路（变形），可以用bellman-ford做（写spfa即可）
判断时判断每个节点被更新的次数，如果超过n次，则说明存在负环回路，退出。。。why？
不明白撒
也可以用floyd来写，这个好写，也好明白，思路跟spfa也差不多
不明白为什么写spfa时候用dis来判断是错的
代码1 floyd
1
#include<stdio.h>
2#include<string.h>
3#include<math.h>
4#define MAX 100
5#define MAXN 50000
6char name[40][1000];
7double f[MAX][MAX];
8int n,m,t;
9void init()
10
{
11
int i,j,a,b;
12 double rat;
13 char s1[1000],s2[1000];
14 for (i=1; i<=n ; i++ )
15
{
16 scanf("%s",&name[i]);
17
}
18 memset(f,1,sizeof(f));
19 scanf("%d",&m);
20 for(i=1; i<=m; i++)
21 {
22 scanf("%s%lf%s",&s1,&rat,&s2);
23 for (j=1; j<=n ; j++ )
24 {
25 if (strcmp(s1,name[j])==0) a=j;
26 if (strcmp(s2,name[j])==0) b=j;
27 }
28 f[a][b]=rat;
29 }
30
}
31void work()
32{
33 int i,j,k;
34 short flag;
35 flag=0;
36 for (k=1; k<=n ; k++ )
37 for (i=1; i<=n ; i++ )
38 for (j=1; j<=n ; j++ )
39 if (f[i][j]<f[i][k]*f[k][j])
40 {
41 f[i][j]=f[i][k]*f[k][j];
42 }
43 for (i=1;i<=n ;i++ )
44 if (f[i][i]>1)
45 {
46 flag=1;
47 break;
48 }
49 if (flag==1)
50 {
51 printf("Case %d: Yes\n",t);
52 }
53 else
54 printf("Case %d: No\n",t);
55}
56int main()
57{
58 t=0;
59 while (scanf("%d",&n)!=EOF&&n!=0)
60 {
61 t++;
62 init();
63 work();
64 }
65 return 0;
66}
67
代码2 spfa 负权回路 1#include<stdio.h>
2#include<string.h>
3#include<math.h>
4#define MAX 10000
5#define MAXN 50000
6char name[35][1000];
7int n,m;
8int e[MAX+5],next[MAX+5],link[MAX+5];
9double rate[MAX+5],dis[MAX+5];
10short vis[MAX+5];
11double total;
12int t,s;
13int g[MAX+1];
14int queue[MAXN+1],head,tail;
15void add(int x,int y,double rat1)
16{
17 s++;
18 e[s]=y;
19 rate[s]=rat1;
20 next[s]=link[x];
21 link[x]=s;
22}
23void init()
24{
25 int i,j;
26 char s1[1000],s2[1000];
27 int a,b;
28 double rn;
29 for (i=1; i<=n ; i++ )
30 {
31 scanf("%s",&name[i]);
32 }
33 s=0;
34 memset(next,0,sizeof(next));
35 memset(link,0,sizeof(link));
36 for (i=1;i<=n ;i++ )
37 {
38 add(0,i,1);
39 }
40 scanf("%d",&m);
41 for(i=1; i<=m; i++)
42 {
43 scanf("%s%lf%s",&s1,&rn,&s2);
44 for (j=1; j<=n; j++)
45 if (strcmp(s1,name[j])==0) a=j;
46 for (j=1; j<=n; j++)
47 if (strcmp(s2,name[j])==0) b=j;
48 add(a,b,rn);
49 }
50}
51int spfa()
52{
53 int i,u,j;
54 memset(vis,0,sizeof(vis));
55 memset(g,0,sizeof(g));
56 queue[1]=0;
57 for (i=1; i<=n; i++)
58 {
59 dis[i]=0;
60 }
61 dis[0]=1;
62 vis[0]=1;
63 head=0;
64 tail=1;
65 while (head!=tail)
66 {
67 head++;
68 u=queue[head];
69 j=link[u];
70 while (j!=0)
71 {
72 if (dis[u]*rate[j]>dis[e[j]])
73 {
74 dis[e[j]]=dis[u]*rate[j];
75 g[e[j]]++;
76 if (g[e[j]]>n)
77 {
78 return 1;
79 }
80 if (!vis[e[j]])
81 {
82 vis[e[j]]=1;
83 tail++;
84 queue[tail]=e[j];
85 }
86 }
87 j=next[j];
88 }
89 vis[u]=0;
90 }
91 return 0;
92}
93void work()
94{
95 int i,flag;
96 flag=spfa();
97 if (flag==1)
98 {
99 printf("Case %d: Yes\n",t);
100 }
101 else
102 printf("Case %d: No\n",t);
103}
104int main()
105{
106 t=0;
107 while (scanf("%d",&n)!=EOF&&n!=0)
108 {
109 t++;
110 init();
111 work();
112 }
113 return 0;
114}
115
我的代码一向冗杂……