[ACM] HDU 1217 arbitrage (bellman_ford is the shortest path to determine whether a positive loop or floyed exists)

Arbitrage

Problem descriptionarbitrage 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.
 
Inputthe input file will contain in 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.
 
Outputfor each test case, print one line telling whether Arbitrage is possible or not in the format "case: yes" respectively "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

 
Sourceuniversity of Ulm Local contest 1996

Solution:

Method 1:

A currency, after several rounds of exchange, returns to the currency to increase its value. This is "Arbitrage ". The bellman_ford algorithm is used to obtain the maximum profit, instead of short circuit. The bellman algorithm is used to change the value. Determine whether a positive loop exists.

Code:

# Include <iostream> # include <string. h> using namespace STD; const int maxn = 35; double dis [maxn]; string STR [maxn]; int nodenum, edgenum; struct edge {int S, E; double W;} edge [maxn * maxn]; bool bellman_ford (INT start) {for (INT I = 1; I <= nodenum; I ++) dis [I] = 0; // Change 1 dis [start] = 1; bool OK; For (INT I = 1; I <= nodeNum-1; I ++) {OK = 0; For (Int J = 1; j <= edgenum; j ++) {If (DIS [edge [J]. s] * edge [J]. w> dis [edge [J]. e]) // change 2 {dis [edge [J]. E] = dis [edge [J]. S] * edge [J]. W; OK = 1 ;}} if (! OK) break;} For (INT I = 1; I <= edgenum; I ++) if (DIS [edge [I]. s] * edge [I]. w> dis [edge [I]. e]) return true; return false;} int main () {int c = 1; while (CIN> nodenum & nodenum) {for (INT I = 1; I <= nodenum; I ++) CIN> STR [I]; CIN> edgenum; string from, to; double W; For (INT I = 1; I <= edgenum; I ++) {CIN> from> W> to; Int J, K; For (j = 1; j <= nodenum; j ++) if (from = STR [J]) break; For (k = 1; k <= nodenum; k ++) if (to = STR [k]) break; edge [I]. S = J; edge [I]. E = K; edge [I]. W = W;} If (bellman_ford (1) cout <"case" <C ++ <": yes" <Endl; else cout <"case" <C ++ <": No" <Endl;} return 0 ;}

Method 2:

Create a graph and an adjacent matrix to find the shortest short circuit between any two edges. If dis [I] [I]> 1, it indicates that a positive weight loop exists. MAP is used in the code, and the ing between string and integer numbers is

Code:

#include <iostream>#include <map>#include <string.h>using namespace std;const int maxn=40;string str[maxn];double mp[maxn][maxn];int nodeNum,edgeNum;void floyed(){    for(int k=1;k<=nodeNum;k++)        for(int i=1;i<=nodeNum;i++)            for(int j=1;j<=nodeNum;j++)            {                if(mp[i][j]<mp[i][k]*mp[k][j])                    mp[i][j]=mp[i][k]*mp[k][j];            }}int main(){    int c=1;    while(cin>>nodeNum&&nodeNum)    {        map<string,int>st;        for(int i=1;i<=nodeNum;i++)        {            cin>>str[i];            st[str[i]]=i;        }        for(int i=1;i<=nodeNum;i++)            for(int j=1;j<=nodeNum;j++)        {            if(i==j)                mp[i][j]=1;            else                mp[i][j]=0;        }        cin>>edgeNum;        string from,to;double w;        for(int i=1;i<=edgeNum;i++)        {            cin>>from>>w>>to;            mp[st[from]][st[to]]=w;        }        floyed();        bool ok=0;        for(int i=1;i<=nodeNum;i++)            if(mp[i][i]>1)             ok=1;        if(ok)            cout<<"Case "<<c++<<": Yes"<<endl;        else            cout<<"Case "<<c++<<": No"<<endl;    }    return 0;}