poj 2240 汇率 floryd算法变形

Arbitrage

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 19824 Accepted: 8391

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

3USDollarBritishPoundFrenchFranc3USDollar 0.5 BritishPoundBritishPound 10.0 FrenchFrancFrenchFranc 0.21 USDollar3USDollarBritishPoundFrenchFranc6USDollar 0.5 BritishPoundUSDollar 4.9 FrenchFrancBritishPound 10.0 FrenchFrancBritishPound 1.99 USDollarFrenchFranc 0.09 BritishPoundFrenchFranc 0.19 USDollar0

Sample Output

Case 1: YesCase 2: No

Source

Ulm Local 1996

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题目大意：可以简单描述为知道从i到j的汇率，问能不能赚钱

解题思路：问能不能赚钱，就是当i到i初始是 t 时，从i到i的最长路径能不能大于t，如果有这样的情况就是能赚钱

方法就是floryd算法变形  重在理解 注意变形 ，多做题才能熟练掌握～～～

#include <stdio.h>#include <algorithm>#include <iostream>#include <string.h>using namespace std;#define INF 0x3f3f3f3f#define N 110double maps[N][N];  /// double 存汇率int n,m;void init(){    int i,j;    for(i=0;i<=n;i++)    {        for(j=0;j<=n;j++)        {            maps[i][j]=(i==j)? 1: 0;         }    }}void floyd(){    int i,j,k;    for(k=1;k<=n;k++)    {        for(i=1;i<=n;i++)        {            for(j=1;j<=n;j++)            {               if(maps[i][k]*maps[k][j]>maps[i][j])  ///若汇率之积大于原来的汇率 则变换                  maps[i][j]=maps[i][k]*maps[k][j];            }        }    }}char str[50][110];char a[110],b[110];int main(){    int i,j,ca=1,k;    double x;    while(scanf("%d",&n),n)    {        init();        for(i=1;i<=n;i++)        {            scanf("%s",str[i]);        }        scanf("%d",&m);        for(i=1;i<=m;i++)        {            scanf("%s %lf %s",a,&x,b);            for(j=1;j<=n;j++)            {                if(strcmp(str[j],a)==0)  ///查找a在str中的位置                    break;            }            for(k=1;k<=n;k++)            {                if(strcmp(str[k],b)==0)  ///查找b的位置                    break;            }            maps[j][k]=x;  ///类似最短路 路径权值～        }        floyd();        printf("Case %d: ",ca++);        for(i=1;i<=n;i++)        {            if(maps[i][i]>1)                break;           ///只要赚钱就跳出即可。。。        }        if(i>n)            printf("No\n");        else            printf("Yes\n");    }    return 0;}

结构体

bellman_ford

：

• Source Code

  #include <iostream>  #include <string.h>  using namespace std;  #define MAXC 100  #define MAXV 50  #define MAXM 10000    typedef struct{      int a,b;      double rate;  }Edge;    Edge edge[MAXM];  int n,m;    void Input(){      char s[MAXV][MAXC],a[MAXC],b[MAXC];      int i,k,j;      for(i=1;i<=n;i++) scanf("%s",s[i]);      scanf("%d\n",&m);      for(i=1;i<=m;i++){          scanf("%s %lf %s",a,&edge[i].rate,b);          for(j=1;j<=n;j++)              if(!strcmp(s[j],a)) break;          for(k=1;k<=n;k++)              if(!strcmp(s[k],b)) break;          edge[i].a=j;          edge[i].b=k;      }  }    int bellman_ford(){      int i,j;      double d[MAXV];      for(i=1;i<=n;i++) d[i]=0;      d[1]=1;      for(i=1;i<=n;i++){          for(j=1;j<=m;j++)              if(d[edge[j].a]*edge[j].rate>d[edge[j].b])                  d[edge[j].b]=d[edge[j].a]*edge[j].rate;      }      for(j=1;j<=m;j++)          if(d[edge[j].a]*edge[j].rate>d[edge[j].b])              return 1;      return 0;  }    int main(){      int cas=1,i;      while(scanf("%d\n",&n) && n){          Input();          printf("Case %d: ",cas++);          if(!bellman_ford()) printf("No\n");          else printf("Yes\n");      }      return 0;  }