POJ - 2240 Arbitrage Floyd_election time limit: 1000ms memory limit: 65536k t-优快云博客

本文链接：https://blog.youkuaiyun.com/JayACM/article/details/78618841

本文介绍了一种通过货币兑换实现财富增长的可能性判断方法。利用Floyd算法，对给定的货币种类及其之间的汇率进行分析，判断是否存在一种货币转换路径使得最终能够以大于初始金额的货币形式返回。

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Arbitrage

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 24944		Accepted: 10568

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

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题意：给出n种货币，以及m种货币之间汇率，问能否通过货币兑换，使得财富增加。

Ac code:

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <stack>
#include <string>
#include <map>
using namespace std;
#define LL long long
#define INF 0x3f3f3f3f
#define fi first
#define se second
#define eps 1
const int mod=1000000000+7;
const int maxn=100+5;

double dis[maxn][maxn];
map<string,int> con;

void Floyd(int n){
    for (int k = 1; k <= n; k++) {
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                dis[i][j] = max(dis[i][j], dis[i][k] * dis[k][j]);
            }
        }
    }
}

void init(int n){
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            if(i!=j) dis[i][j]=0;
            else dis[i][j]=1;
        }
    }
}

int main() {
    int n,cnt=0;
    while(~scanf("%d",&n)&&n){
        cnt++;
        init(n);
        for(int i=1;i<=n;i++){
            string s;
            cin>>s;
            con[s]=i;
        }
        int m;
        scanf("%d",&m);
        for(int i=1;i<=m;i++){
            string s1,s2;
            double c;
            cin>>s1>>c>>s2;
            dis[con[s1]][con[s2]]=c;
        }
        Floyd(n);
        int flag=0;
        for(int i=1;i<=n;i++){
            if(dis[i][i]>1){
                flag=1;
                break;
            }
        }
        printf("Case %d: ",cnt);
        if(flag) printf("Yes\n");
        else printf("No\n");
    }

}