poj 2240__Arbitrage (Floyd)

Arbitrage

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18152		Accepted: 7676

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

 

鏈接：http://poj.org/problem?id=2240

 

題意：給出貨幣間的匯率，問能否交換貨幣之後獲利。

 

想法1：首先想到dfs，每次從一個貨幣開始搜索，設ans=1，每次把乘上貨幣的匯率，直到找到起始貨幣，如果ans>1，標記之後回溯，每次回溯的時候把ans還原。

結果：TLE。

代碼如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
int n,m,s,flag,vis[33][33];
double rt,ch[33][33],ans;
char ss[110],a[110],b[110];
map<string,int> ma;
void dfs(int aa)
{
    double ans1;
    for(int i=0;i<n;i++) {
        if(ch[aa][i]&&vis[aa][i]==0) {
            vis[aa][i]=1;
            ans1=ans;
            ans*=ch[aa][i];
            if(i==s&&ans>1) {
                flag=1;return ;
            }
            dfs(i);
            if(flag)
                return ;
            ans=ans1;
            vis[aa][i]=0;
        }
    }
}
int main()
{
    int k=1;
    while(~scanf("%d",&n)&&n) {
        for(int i=0;i<n;i++) {	把貨幣名稱映射爲數字，便於使用鄰接矩陣。
            scanf("%s",ss);
            ma[ss]=i;
        }
        scanf("%d",&m);
        for(int i=0;i<m;i++) {
            scanf("%s%lf%s",a,&rt,b);
            ch[ma[a]][ma[b]]=rt;
        }
        flag=0;
        memset(vis,0,sizeof(vis));
        for(s=0;s<n;s++) {
            ans=1.0;

            dfs(s);
        }
        if(flag==1) {
            printf("Case %d: Yes\n",k++);
        }
        else
            printf("Case %d: No\n",k++);
    }
    return 0;
}

想法2：問題的實質相當於求有環最長路徑，用floyd處理是不錯的選擇。
結果：AC。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
int n,m,k=1;
double ch[33][33];
int main()
{
    double rt;
    map<string,int> ma;
    char ss[110],a[110],b[110];
    while(~scanf("%d",&n)&&n) {
        memset(ch,0,sizeof(ch));
        for(int i=0;i<n;i++) {
            scanf("%s",&ss);
            ch[i][i]=1;
            ma[ss]=i;
        }
        scanf("%d",&m);
        for(int i=0;i<m;i++) {
            scanf("%s%lf%s",a,&rt,b);
            ch[ma[a]][ma[b]]=rt;
        }
        for(int i=0;i<n;i++) {
            for(int j=0;j<n;j++) {
                for(int k=0;k<n;k++) {
                    if(ch[i][j]<ch[i][k]*ch[k][j]) {
                        ch[i][j]=ch[i][k]*ch[k][j];
                    }
                }
            }
        }
        int flag=0;
        for(int i=0;i<n;i++) {
            if(ch[i][i]>1) {
                flag=1;
                break;
            }
        }
        if(flag)
            printf("Case %d: Yes\n",k++);
        else
            printf("Case %d: No\n",k++);
    }
    return 0;
}