描述
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.
- 输入
- The input file 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. - 输出
- For each test case, print one line telling whether arbitrage is possible or not in the format "Case case: Yes" respectively "Case case: No".
- 样例输入
-
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
- 样例输出
-
Case 1: Yes Case 2: No
- 来源
- NKOJ or 1996/97 Ulm Internal Contest
- 上传者
- 苗栋栋
题意：给出一些货币和货币之间的兑换比率，问是否可以使某种货币经过一些列兑换之后，货币值增加。举例说就是1美元经过一些兑换之后，超过1美元。可以输出Yes，否则输出No。
AC代码:
1 #include <vector> 2 #include <map> 3 #include <set> 4 #include <algorithm> 5 #include <iostream> 6 #include <cstdio> 7 #include <cmath> 8 #include <cstdlib> 9 #include <string> 10 #include <cstring> 11 #include <queue> 12 using namespace std; 13 #define INF 0x3f3f3f3f 14 #define MAX 111 15 16 double mp[MAX][MAX]; 17 int n,m; 18 19 void floyd() 20 { 21 for(int k=1; k<=n; k++) 22 for(int i=1; i<=n; i++) 23 for(int j=1; j<=n; j++) 24 if(mp[i][j]< mp[i][k]*mp[k][j]) 25 mp[i][j]=mp[i][k]*mp[k][j]; 26 } 27 28 void init() 29 { 30 for(int i=1; i<=n; i++){ 31 for(int j=1; j<=n; j++){ 32 if(i==j) 33 mp[i][j]=1; 34 else 35 mp[i][j]=0; 36 } 37 } 38 } 39 40 int main() 41 { 42 int sum=0; 43 double rate; 44 char a[111],b[111],c[111]; 45 while(~scanf("%d",&n)&&n){ 46 init(); 47 map<string,int> mmp; 48 for(int i=1; i<=n; i++){ 49 scanf("%s",a); 50 mmp[a]=i; 51 } 52 scanf("%d",&m); 53 for(int i=1; i<=m; i++){ 54 scanf("%s%lf%s",b,&rate,&c); 55 int x=mmp[b]; 56 int y=mmp[c]; 57 mp[x][y]=rate; 58 //printf("%d ",mp[x][y]); 59 } 60 floyd(); 61 int flag=0; 62 for(int i=1; i<=n; i++){ 63 //printf("%d ",mp[i][i]); 64 if(mp[i][i]>1){ 65 flag=1; 66 break; 67 } 68 } 69 printf("Case %d: ",++sum); 70 printf("%s ",flag ? "Yes" : "No"); 71 } 72 }
SPFA:
1 #include <vector> 2 #include <map> 3 #include <set> 4 #include <algorithm> 5 #include <iostream> 6 #include <cstdio> 7 #include <cmath> 8 #include <cstdlib> 9 #include <string> 10 #include <cstring> 11 #include <queue> 12 using namespace std; 13 #define INF 0x3f3f3f3f 14 #define MAX 111 15 16 int n, m; 17 double dis[MAX], mp[MAX][MAX]; 18 struct node 19 { 20 char name[100]; 21 }a[MAX]; 22 23 int find(char *s) 24 { 25 for(int i = 0; i < n; i++) 26 if(strcmp(a[i].name, s) == 0) 27 return i; 28 } 29 30 int SPFA(int p) 31 { 32 queue<int> q; 33 bool vis[MAX]; 34 memset(dis,0,sizeof(dis)); 35 memset(vis, 0, sizeof(vis)); 36 while(!q.empty()) 37 q.pop(); 38 dis[p] = 1; 39 vis[p] = 1; 40 q.push(p); 41 while(!q.empty()) 42 { 43 int x = q.front(); 44 q.pop(); 45 vis[x] = false; 46 for(int i = 0; i < n; i++) 47 { 48 if(dis[i] < dis[x] * mp[x][i]) 49 { 50 dis[i] = dis[x] * mp[x][i]; 51 if(dis[p] > 1.0) 52 return 1; 53 if(!vis[i]) 54 { 55 vis[i] = true; 56 q.push(i); 57 } 58 } 59 } 60 } 61 return 0; 62 } 63 64 int main() 65 { 66 int i, j, cas = 0; 67 char s1[35], s2[35]; 68 double s; 69 while(~scanf("%d",&n) && n) 70 { 71 for(i = 0; i < n; i++) 72 { 73 for(j = 0; j < n; j++) 74 { 75 if(i == j) 76 mp[i][j] = 1; 77 else 78 mp[i][j] = 0; 79 } 80 } 81 for(i = 0; i < n; i++) 82 scanf("%s",a[i].name); 83 scanf("%d",&m); 84 for(i = 0; i < m; i++) 85 { 86 scanf("%s%lf%s",s1, &s, s2); 87 int u = find(s1), v = find(s2); 88 mp[u][v] = s; 89 } 90 int flag = 0; 91 for(i = 0; i < n; i++) 92 { 93 if(SPFA(i) == 1) 94 { 95 flag = 1; 96 break; 97 } 98 } 99 printf("Case %d: ",++cas); 100 printf("%s ", flag ? "Yes" : "No"); 101 } 102 return 0; 103 }
Bellman_Ford代码(hdu 可过):
1 #include <vector> 2 #include <map> 3 #include <set> 4 #include <algorithm> 5 #include <iostream> 6 #include <cstdio> 7 #include <cmath> 8 #include <cstdlib> 9 #include <string> 10 #include <cstring> 11 #include <queue> 12 using namespace std; 13 #define INF 0x3f3f3f3f 14 #define MAX 111 15 16 struct node 17 { 18 int x,y; 19 double rate; 20 }e[MAX]; 21 22 int n,m,v; 23 bool flag; 24 double dis[MAX]; 25 26 bool Bellman_Ford(int p) 27 { 28 memset(dis,0,sizeof(dis)); 29 dis[p]=1; 30 for(int j=0; j<n; j++) 31 for(int i=0; i<v; i++) 32 { 33 if(dis[e[i].y] < dis[e[i].x] * e[i].rate) 34 dis[e[i].y] = dis[e[i].x] * e[i].rate; 35 } 36 //for(int i=0; i<v; i++) 37 // printf("%d ",dis[e[i].y]); 38 for(int i = 0; i<v; i++) 39 if(dis[e[i].y] < dis[e[i].x] * e[i].rate) 40 return true; 41 return false; 42 } 43 44 int main() 45 { 46 int sum=0; 47 char a[111], b[111], c[111]; 48 double rate; 49 while(~scanf("%d",&n)&&n){ 50 v=0; 51 map<string,int> mp; 52 for(int i=1; i<=n; i++){ 53 scanf("%s",a); 54 mp[a]=i; 55 } 56 scanf("%d",&m); 57 for(int i=1; i<=m; i++){ 58 scanf("%s%lf%s",b,&rate,c); 59 int x=mp[b]; 60 int y=mp[c]; 61 e[v].x=x; 62 e[v].y=y; 63 e[v++].rate=rate; 64 } 65 flag=Bellman_Ford(1); 66 if (flag) 67 printf("Case %d: Yes ",++sum); 68 else 69 printf("Case %d: No ", ++sum); 70 } 71 }