DFS判断正环

最新推荐文章于 2022-07-07 23:37:57 发布

转载最新推荐文章于 2022-07-07 23:37:57 发布 · 84 阅读

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原文链接：http://www.cnblogs.com/mypsq/p/4348232.html

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#java

本文介绍了一种通过检测货币兑换网络中的正环来判断是否存在获利机会的方法。利用图论中的负权边算法，通过示例详细说明了如何构建图模型，并实现了一个能够检测是否存在获利可能的程序。

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hdu1217

Arbitrage

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4123 Accepted Submission(s): 1878

Problem 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 file 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
  

程序：

#include"stdio.h"
#include"string.h"
#include"iostream"
#include"map"
#include"string"
#include"queue"
#include"stdlib.h"
#include"math.h"
#define M 40
#define eps 1e-10
#define inf 99999999
#define mod 1000000000
using namespace std;
struct st
{
    int u,v,next;
    double w;
}edge[M*M*2];
int head[M],t,cnt[M],n;
double dis[M];
void init()
{
    t=0;
    memset(head,-1,sizeof(head));
}
void add(int u,int v,double w)
{
    edge[t].u=u;
    edge[t].v=v;
    edge[t].w=w;
    edge[t].next=head[u];
    head[u]=t++;
}
int dfs(int u)
{
    cnt[u]=1;
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(dis[v]<dis[u]*edge[i].w)
        {
            dis[v]=dis[u]*edge[i].w;
            if(cnt[v])
                return 1;
            if(dfs(v))
                return 1;
        }
    }
    cnt[u]=0;
    return 0;
}
int solve()
{
     memset(cnt,0,sizeof(cnt));
     for(int i=1;i<=n;i++)//防止题目不连通的情况
     {
          memset(dis,0,sizeof(dis));
          dis[i]=1;
          if(dfs(i))
               return 1;
     }
     return 0;
}
int main()
{
    int m,i,kk=1;
    while(scanf("%d",&n),n)
    {
        map<string,int>mp;
        char ch1[111],ch2[111];
        for(i=1;i<=n;i++)
        {
            scanf("%s",ch1);
            mp[ch1]=i;
        }
        scanf("%d",&m);
        init();
        double c;
        while(m--)
        {
            scanf("%s%lf%s",ch1,&c,ch2);
            add(mp[ch1],mp[ch2],c);
        }
        int ans=solve();
        printf("Case %d: ",kk++);
        if(ans)
            printf("Yes\n");
        else
            printf("No\n");
    }
}

转载于:https://www.cnblogs.com/mypsq/p/4348232.html