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poj 2240 Arbitrage（Bellman-Ford算法学习）

2024-07-18 23:46:20   224人阅读

Arbitrage
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 15806 Accepted: 6648

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

3
USDollar
BritishPound
FrenchFranc
3
USDollar 0.5 BritishPound
BritishPound 10.0 FrenchFranc
FrenchFranc 0.21 USDollar

3
USDollar
BritishPound
FrenchFranc
6
USDollar 0.5 BritishPound
USDollar 4.9 FrenchFranc
BritishPound 10.0 FrenchFranc
BritishPound 1.99 USDollar
FrenchFranc 0.09 BritishPound
FrenchFranc 0.19 USDollar

0

Sample Output

Case 1: Yes
Case 2: No

Source

Ulm Local 1996

算法思想：从源点逐次途经其他顶点，以缩短到达终点的最短路径长度。限制：图中不含权值总和为负值的回路。


#include"stdio.h"
#include"string.h"
#include"queue"
#include"iostream"
#include"algorithm"
using namespace std;
#define N 50
#define LL __int64
const int inf=0;
struct node
{
    int u,v;
    double d;
}e[N*N];
int m,n;
double dis[N];
bool bellman()
{
    int i,j,u,v;
    for(i=0;i<n;i++)      //初始化为0
        dis[i]=inf;
    dis[0]=1;
    for(i=1;i<=n;i++)     //从dis[0]递推到dis[1]....dis[n];
    {
        for(j=0;j<m;j++)
        {
            u=e[j].u;
            v=e[j].v;
            if(dis[u]&&dis[v]<dis[u]*e[j].d)   //若经过某一点能使从源点到该终点的最短路缩短，此为增大
                dis[v]=dis[u]*e[j].d;          //则修改该点值，即松弛
        }
    }
    return dis[0]>1;
}
int main()
{
    int i,j,k,cnt=1;
    double d;
    char str[N][N],a[N],b[N];
    while(scanf("%d",&n),n)
    {
        for(i=0;i<n;i++)
            scanf("%s",str[i]);
        scanf("%d",&m);
        for(i=0;i<m;i++)
        {
            scanf("%s %lf %s",a,&d,b);
            for(j=0;strcmp(str[j],a);j++)
                ;
            for(k=0;strcmp(str[k],b);k++)
                ;
            e[i].u=j;
            e[i].v=k;
            e[i].d=d;
        }
        printf("Case %d: ",cnt++);
        bool flag=bellman();
        if(flag)
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}


poj 2240 Arbitrage（Bellman-Ford算法学习）


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