POJ 2240 Arbitrage / ZOJ 1092 Arbitrage / HDU 1217 Arbitrage / SPOJ Arbitrage(图论,环)

POJ 2240 Arbitrage / ZOJ 1092 Arbitrage / HDU 1217 Arbitrage / SPOJ Arbitrage(图论,环)

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

Http

POJ:https://vjudge.net/problem/POJ-2240
ZOJ:https://vjudge.net/problem/ZOJ-1092
HDU:https://vjudge.net/problem/HDU-1217
SPOJ:https://vjudge.net/problem/SPOJ-ARBITRAG

Source

图论,环

题目大意

给出m对n种货币的兑换规则,问能否通过这个规则得到更多的钱

解决思路

还是用spfa求是否有环,注意因为本题是有向图,所以不一定连通,要一个一个判

代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<map>
#include<string>
#include<queue>
#include<vector>
using namespace std;

const int maxN=40;
const int inf=2147483647;

class Data
{
public:
    int v;
    double w;
};

int n,m;
vector<Data> E[maxN];
map<string,int> Map;
queue<int> Q;
bool inqueue[maxN];
double Dist[maxN];

bool spfa(int s);

int main()
{
    int ti=0;
    while (cin>>n)
    {
        if (n==0)
            break;
        Map.clear();
        for (int i=1;i<=n;i++)
            E[i].clear();
        for (int i=1;i<=n;i++)
        {
            string str;
            cin>>str;
            Map[str]=i;
        }
        cin>>m;
        for (int i=1;i<=m;i++)
        {
            string str1,str2;
            double r;
            cin>>str1>>r>>str2;
            E[Map[str1]].push_back((Data){Map[str2],r});
        }
        bool is_ans=0;
        for (int i=1;i<=n;i++)
            if (spfa(i)==1)
            {
                is_ans=1;
                break;
            }
        ti++;
        if (is_ans==1)
            printf("Case %d: Yes\n",ti);
        else
            printf("Case %d: No\n",ti);
    }
    return 0;
}

bool spfa(int s)
{
    memset(Dist,0,sizeof(Dist));
    memset(inqueue,0,sizeof(inqueue));
    while (!Q.empty())
        Q.pop();
    Dist[s]=1;
    Q.push(s);
    inqueue[s]=1;
    do
    {
        if (Dist[s]>1)
            return 1;
        int u=Q.front();
        Q.pop();
        inqueue[u]=0;
        for (int i=0;i<E[u].size();i++)
        {
            int v=E[u][i].v;
            if (Dist[v]<Dist[u]*E[u][i].w)
            {
                Dist[v]=Dist[u]*E[u][i].w;
                if (inqueue[v]==0)
                {
                    Q.push(v);
                    inqueue[v]=1;
                }
            }
        }
    }
    while (!Q.empty());
    return 0;
}
posted @ 2017-07-24 11:25 SYCstudio 阅读(...) 评论(...) 编辑 收藏