hdu 1217 利用flord算法求 有环图 2点之间最大值                                           Arbitrage

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.

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.

For each test case, print one line telling whether arbitrage is possible or not in the format "Case case: Yes" respectively "Case case: No".

University of Ulm Local Contest 1996

Eddy

 

题意   ：输入n  表示有n中货币，输入m   之后m行 每行格式为  货币名 兑换比率  货币名 即  a   p   b  表示1单元货币a可以兑换p单位b  

问这样兑换形成一个圈之后回到原来的货币，能不能赚到钱   即用1单元的货币和其他货币兑换 若干次兑换之后兑换回来原来的货币 是不是货币量大于1了

是则输出Yes

 

思路：

map+最短路（flord算法)

利用了flord的性质求最大值  很帅

具体看代码

：

#include<stdio.h>
#include<map>
#include<string>
#include<string.h>
using namespace std;
double a[100][100];
char s1[1000],s2[1000];
int pos,n,flag,time,vis[50];
double mmax(double a,double b)
{
    if(a>b) return a;
    else return b;
}
int main()
{
      int m,i,j,k,cas=0;
      char s[1000];
      while(scanf("%d",&n)!=EOF)
      {
          if(!n) break;
          cas++;
          map<string,int>mp;
          int cnt=0;
          for(i=0;i<n;i++)
          {
              scanf("%s",s);
              if(mp.find(s)==mp.end()) mp[s]=cnt++;
          }
          for(i=0;i<n;i++)
              for(j=0;j<n;j++)
                  a[i][j]=0.0;
          scanf("%d",&m);
          while(m--)
          {
              int  id1,id2;
              double mid;
                 scanf("%s %lf %s",s1,&mid,s2);
                 if(mp.find(s1)==mp.end()||mp.find(s2)==mp.end()) continue;
                 id1=mp[s1];
                 id2=mp[s2];

                 a[id1][id2]=mid;
                // printf("a[%d][%d]=%lf
",id1,id2,mid);
          }
        flag=0;
        for(int k=0;k!=n;k ++)
           for(int i=0;i!=n;i ++)
              for(int j=0;j!=n; j ++)
               {
                     a[i][j]=mmax(a[i][j],a[i][k]*a[k][j]);
                     if(a[i][j]> 1 && i==j)
                      {
                             flag=true;
                               break;
                       }
                }

        if(flag) printf("Case %d: Yes
",cas);
        else printf("Case %d: No
",cas);
      }
      return 0;
}