ACM: 图论题 poj 2240 (floyd + ST…-优快云博客

Arbitrage

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

USDollarBritish

PoundFrench

Franc

USDollar 0.5 BritishPound

BritishPound 10.0 FrenchFranc

FrenchFranc 0.21 USDollar

USDollar

BritishPound

FrenchFranc

USDollar 0.5 BritishPound

USDollar 4.9 FrenchFranc

BritishPound 10.0 French

FrancBritishPound 1.99 USDollar

FrenchFranc 0.09 BritishPound

FrenchFranc 0.19 USDollar

Sample Output

Case 1: Yes

Case 2: No

题意: 判断每种货币结果货币转换是否的赚钱. (汇率结果转换后大于1.0,炒货币一个意思)

解题思路:
1. 明显的最短路径变形题.(每种货币都要判断一次),果断floyd算法.
2. 思路的感觉非常重要一次AC.

代码:

#include <cstdio>
#include <iostream>
#include <string>
#include <cstdlib>
#include <map>
using namespace std;
#define MAX 100
const int INF = (1<<20);
typedef map<string,int> UDT_MAP_STR_INT;

int n , m;
double edges[MAX][MAX];
UDT_MAP_STR_INT str_int_map;

void read_graph()
{
   char str1[MAX];
   char str2[MAX];
   int i , j;
   double ww;
   str_int_map.clear();
   for(i = 1; i <= n; ++i)
   {
       scanf("%s",str1);
       str_int_map.insert(map<string,int>::value_type((string)str1,i));
       edges[i][i] = 1;
   }

   scanf("%d",&m);
   for(i = 1; i <= m; ++i)
   {
       scanf("%s %lf %s",str1,&ww,str2);
       edges[ str_int_map[(string)str1] ][ str_int_map[(string)str2] ] = ww;
   }
}

bool floyd()
{
   for(int k = 1; k <= n; ++k)
   {
       for(int i = 1; i <= n; ++i)
       {
           for(int j = 1; j <= n; ++j)
           {
               if(edges[i][j] < edges[i][k] * edges[k][j])
               {
                   edges[i][j] = edges[i][k] * edges[k][j];
               }
           }
       }
   }

   for(int p = 1; p <= n; ++p)
   {
       if(edges[p][p] > 1)
           return true;
   }
   return false;
}

int main()
{
   freopen("input.txt","r",stdin);
   int num = 1;
   while(scanf("%d",&n) != EOF && n != 0)
   {
       read_graph();
       if(floyd() == true)
           printf("Case %d: Yes\n",num++);
       else
           printf("Case %d: No\n",num++);
   }

   return 0;
}