poj 2240 Arbitrage

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Time Limit: 1000 MS Memory Limit: 65536 KB

64-bit integer IO format: %I64d , %I64u   Java class name: Main

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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 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. OutputFor 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
題意：對於每一個頂點構成的迴路  兌換率是否大於1

 #include <iostream> #include <string.h> #include <stdio.h> using namespace std; #define maxx 35 #define maxn 1000 char name[maxx][20],a[20],b[20];  ///鏈表進行儲存 double maxdis[maxn];  ///類似於dis[]數組  不過這次求最大迴路 double x; ///兌換率 int n,t; int falg; struct exchange {     int ci,cj;     double cij; }ex[maxn]; void Bellman(int v0) {     falg=0;     memset(maxdis,0,sizeof(maxdis));     maxdis[v0]=1;     for(int k=1;k<=n;k++)  ///要尋找迴路  所以從maxdis[0]遞推maxdis[1]....maxdis[n]     {         for(int i=0;i<t;i++)  ///每條邊加入是否值變大使最大         {             if(maxdis[ex[i].ci]*ex[i].cij>maxdis[ex[i].cj])  ///是程的關係了~~~             {                 maxdis[ex[i].cj]=maxdis[ex[i].ci]*ex[i].cij;  ///求最大的  遇到大的就更新             }         }     }     if(maxdis[v0]>1)     falg=1; } int main() {     int num;     int casee=0;     while(scanf("%d",&n),n)     {         for(int i=0;i<n;i++)         {             cin>>name[i];             //scanf("%s",name[num]);         }         int i,j,k;         scanf("%d",&t);         for(i=0;i<t;i++)         {             cin>>a>>x>>b;             //scanf("%s%if%s",a,&x,b);             for(j=0;strcmp(a,name[j]);j++); ///，是空迴圈  沒有迴圈語句的意思             for(k=0;strcmp(b,name[k]);k++);  ///將字串變成了數字  對應關係             ex[i].ci=j;             ex[i].cij=x;             ex[i].cj=k;             ///cout<<i<<‘ ‘<<j<<‘ ‘<<k<<"!!!!!!!"<<endl;  輸出一次啊   挺神奇的東東         }         for(int i=0;i<n;i++)         {             Bellman(i);  ///遍曆每一個點的迴路             if(falg)  ///如果出現兌換率〉1的現象  標記直接退出就好             break;         }         if(falg)         printf("Case %d: Yes\n",++casee);         else         printf("Case %d: No\n",++casee);     }     return 0; }

 

poj 2240 Arbitrage