POJ 2240 Arbitrage

Topic Links:

http://poj.org/problem?id=2240

Title Description:

To the N currencies, to the exchange rate between M-currencies, ask whether the conversion between the money to gain benefits,

Problem Solving Ideas:

By test instructions know, this problem not only find a path, so the most suitable is to choose Floyd to solve the problem, Judge Map[i][i] there is no value greater than 1

Ps:floyd template title.

1#include <cstdio>2#include <cstring>3#include <cstdlib>4#include <algorithm>5#include <iostream>6 using namespacestd;7 #defineMAXN 358 CharNAME[MAXN][MAXN];//the names of various currencies9 DoubleMAP[MAXN][MAXN];//Map[i][j]i-->j's exchange rateTen intN; One  A voidinit (); - intFind (Charstr[]); - voidFloyd (); the  - intMain () - { -     intM, I, j, k =0; +     Doubles; -     CharSTR1[MAXN], STR2[MAXN]; +  A      while(SCANF ("%d", &N), N) at     { - init (); -          for(i=0; i<n; i++) -scanf ("%s", Name[i]); -scanf ("%d", &m); -          while(M--) in         { -scanf ("%s%lf%s", STR1, &s, str2); to             intA =find (STR1); +             intb =find (STR2); -MAP[A][B] =s; the         } * Floyd (); $m =0;Panax Notoginseng          for(i=0; i<n; i++) -                 if(Map[i][i] >1) them =1; +         if(m) Aprintf ("Case %d:yes\n", ++k); the         Else +printf ("Case %d:no\n", ++k); -     } $     return 0; $ } -  - voidInit ()//initialized to a the { -     intI, J;Wuyi      for(i=0; i<n; i++) the          for(j=0; j<n; J + +) -MAP[I][J] =1; Wu } - intFind (Charstr[]) About { $     inti; -      for(i=0; i<n; i++) -         if(!strcmp (name[i], str)) -             returni; A } + voidFloyd () the { -     intI, J, K; $      for(i=0; i<n; i++) the          for(k=0; k<n; k++) the              for(j=0; j<n; J + +) theMAP[I][J] = max (Map[i][j], map[i][k]*map[k][j]); the}

POJ 2240 Arbitrage