Arbitrage

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 30790		Accepted: 12761

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.

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.

For each test case, print one line telling whether arbitrage is possible or not in the format "Case case: Yes" respectively "Case case: No".

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

Case 1: Yes

Case 2: No

#include <stdio.h>

#include <memory.h>

#include <string.h>

#include <queue>

#define MAXN 31

int n,m,head[MAXN],vis[MAXN],cnt[MAXN];

char name[][];　　//用来读入各种货币的名称

using namespace std;

typedef struct

{

    int start,end;

    int next;

    double w;

} Edge;

Edge edges[];

double dis[MAXN];

int Find(char str1[])

{

    for(int i=; i<=n; i++)

        if(strcmp(name[i],str1)==)

            return i;//没有考虑找不到的情况，但是实际上，题目不会给出找不到的货币名称....

    return -;

}

bool spaf(int s)

{

    queue<int>que;

    memset(dis,,sizeof(dis));

    memset(vis,false,sizeof(vis));

    memset(cnt,,sizeof(cnt));

    vis[s]=dis[s]=;　　　　//这里一开始学习spfa的时候，还保留着用其他方法的习惯，总给vis[1]置为1 ，，但其实这里应该写成vis[s]=1，不然题目就会出现一些小错误，导致过了样例，却还是wa了

    que.push(s);

    while(!que.empty())

    {

        int temp=que.front();

        que.pop();

        vis[temp]=;

        for(int i=head[temp]; i!=-; i=edges[i].next)

        {int x=edges[i].start,y=edges[i].end;

            double w=edges[i].w;

            if(dis[y]<dis[x]*w)

            {

                dis[y]=dis[x]*w;

                if(!vis[y])

                {

                    vis[y]=;

                    if(++cnt[y]>n)　　　　//为了防止某一个节点无限被调用，形成环路，则可以一直赚钱，卡在这里

                        return true;

                    que.push(y);

                }

            }

        }

    }

    return false;

}

int main()

{

    int len=;

    char start[],end[];

    while(scanf("%d",&n)&&n!=)

    {

        getchar();

        for(int i=; i<=n; ++i)

            gets(name[i]);

        scanf("%d",&m);

        memset(head,-,sizeof(head));

        for(int i=; i<m; ++i)

        {

            scanf(" %s %lf %s",start,&edges[i].w,end);

            edges[i].start=Find(start);

            edges[i].end=Find(end);

            edges[i].next=head[edges[i].start];

            head[edges[i].start]=i;

        }

        m=;

        for(int i=; i<=n; ++i)

        {

            if(spaf(i)==true)　　　　

            {

                m=;

                break;

            }

        }

        if(m)

            printf("Case %d: Yes\n",len++);

        else

            printf("Case %d: No\n",len++);

    }

    return ;

}