HDU 1271 arbitrage

Link:

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1217

Question:

Arbitrage

Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 2594 accepted submission (s): 1167


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

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Analysis and Summary:

According to the meaning of the question, only the positive ring can be determined as yes. Therefore, spfa can determine the ring, or after the adjacent matrix graph is created and calculated using the Floyd algorithm, determine whether d [I] [I]> 1 is available.

1. Floyd

#include<cstdio>#include<iostream>#include<cstring>#include<queue>#include<map>#include<string>using namespace std;typedef double Type;const int INF = 0x7ffffff;const int VN  = 35;const int EN  = VN*VN;map<string, int>mp;int n,size,cnt;int head[VN];Type d[VN][VN];void init(){    size=0; cnt=0;    mp.clear();    for(int i=1; i<=n; ++i){        d[i][i] = 1;        for(int j=i+1; j<=n; ++j)            d[i][j] = d[j][i] = 0;    }}void Floyd(){    for(int k=1; k<=n; ++k)    for(int i=1; i<=n; ++i)    for(int j=1; j<=n; ++j){        d[i][j] = max(d[i][j], d[i][k]*d[k][j]);    }}int main(){    char str1[50], str2[50];    int cas=1,m,u,v;    double w;    while(scanf("%d",&n)&&n){        init();        for(int i=0; i<n; ++i){            scanf("%s",str1);            mp[str1] = ++cnt;        }        scanf("%d",&m);        for(int i=0; i<m; ++i){            scanf("%s %lf %s",str1,&w,str2);            u = mp[str1], v = mp[str2];            d[u][v] = w;        }        bool flag=false;        Floyd();        for(int i=1; i<=n; ++i)            if(d[i][i]>1.0) {flag=true; break;}        printf("Case %d: ",cas++);        if(flag) puts("Yes");        else puts("No");    }    return 0;}

2. spfa judgment Ring

#include<cstdio>#include<iostream>#include<cstring>#include<string>#include<queue>#include<map>using namespace std;typedef double Type;const int INF = 0x7ffffff;const int VN  = 35;const int EN  = VN*VN;map<string, int>mp;int n,size,cnt;int head[VN];bool inq[VN];int counter[VN];Type d[VN];struct Edge{int v,next; Type w;}E[EN];void addEdge(int u,int v,Type w){    E[size].v=v, E[size].w=w;    E[size].next = head[u];    head[u] = size++;}void init(){    size=0; cnt=0;    memset(head, -1, sizeof(head));    mp.clear();}bool SPFA(int src){    memset(counter, 0, sizeof(counter));    memset(inq, 0, sizeof(inq));    for(int i=1; i<=n; ++i) d[i]=0;    d[src] = 1;    queue<int>q;    q.push(src);    while(!q.empty()){        int u = q.front();  q.pop();        inq[u] = false;        for(int e=head[u]; e!=-1; e=E[e].next){            Type tmp = d[u]*E[e].w;           // if(d[E[e].v] > tmp){            if(d[E[e].v]<tmp){                d[E[e].v] = tmp;                if(!inq[E[e].v]){                    inq[E[e].v] = true;                    q.push(E[e].v);                    if(++counter[E[e].v]>=n){                        return true;                    }                }            }                    }    }    return false;}int main(){    char str1[50], str2[50];    int cas=1,m,u,v;    double w;    while(scanf("%d",&n)&&n){        init();        for(int i=0; i<n; ++i){            scanf("%s",str1);            mp[str1] = ++cnt;        }        scanf("%d",&m);        for(int i=0; i<m; ++i){            scanf("%s %lf %s",str1,&w,str2);            u = mp[str1], v = mp[str2];            addEdge(u,v,w);        }        bool flag=false;        for(int i=1; i<=n; ++i){            if(SPFA(i)){                flag=true;                break;            }        }        printf("Case %d: ",cas++);        if(flag) puts("Yes");        else puts("No");    }    return 0;}

 
  
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   Http://blog.csdn.net/shuangde800
   , By d_double (reprinted, please mark)