hdu 1217 Arbitrage （Floyd算法）

Arbitrage

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3196    Accepted Submission(s): 1453

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

3USDollarBritishPoundFrenchFranc3USDollar 0.5 BritishPoundBritishPound 10.0 FrenchFrancFrenchFranc 0.21 USDollar3USDollarBritishPoundFrenchFranc6USDollar 0.5 BritishPoundUSDollar 4.9 FrenchFrancBritishPound 10.0 FrenchFrancBritishPound 1.99 USDollarFrenchFranc 0.09 BritishPoundFrenchFranc 0.19 USDollar0

 

Sample Output

Case 1: YesCase 2: No

 

Source

University of Ulm Local Contest 1996

 

Recommend

Eddy

解题报告：第一次看到这个题感觉应该是dijkstra，只不过加法改成了乘法嘛，应该很简单的题目：但是提交后愣是wa,看着代码好像没有问题了，在discuss区看看都是用floyd做的，有位大牛说乘法中小于1的数，就相当于加法中的负数，（这个貌似数学老师在哪说过啊，不记得了。。）所以前面果断wa,自己又用Floyd写了一遍果断ac；

代码：

#include<cstdio>#include<cstring>#include<map>#include<string>using namespace std;int n,m;map<string,int>co;double dis[35][35];void floyd(){for(int k=1;k<=n;k++)for(int i=1;i<=n;i++)for(int j=1;j<=n;j++){if(dis[i][j]<dis[i][k]*dis[k][j])dis[i][j]=dis[i][k]*dis[k][j];}}int main(){int ncase = 0;char str[100],str1[100],str2[100];double rate;while(scanf("%d",&n),n){memset(dis,0,sizeof(dis));for(int i=1;i<=n;i++){scanf("%s",str);co[str]=i;}scanf("%d",&m);for(int i=0;i<m;i++){scanf("%s%lf%s",str1,&rate,str2);dis[co[str1]][co[str2]]=rate;}floyd();printf("Case %d: ",++ncase);//printf("%lf\n",dis[1][1]);if(dis[1][1]>1.0) printf("Yes\n");else printf("No\n");} return 0;}

dijkstra的错误方法留作纪念：

#include<cstdio>#include<cstring>#include<map>#include<string>using namespace std;int n,m,flag;const int maxn  = 0x7ffffff;double dis[40][40],len[40],ans;int vis[40];map<string,int>co;double Max(double a,double b){return a>b?a:b;}void dijkstra(int st){memset(vis,0,sizeof(vis));int min,cnt;ans=1.0;for(int i=1;i<=n;i++)len[i]=dis[st][i];for(int i=1;i<=n;i++){double min=0;int cnt=-1;for(int j=1;j<=n;j++ )if(!vis[j]&&min+0.001 < len[j])min=len[cnt=j];if(cnt==-1){ans=0; return ;}ans=ans*min;vis[cnt]=1;for(int j=1;j<=n;j++)if(!vis[j] && dis[cnt][j]!=0.000000)len[j]=dis[cnt][j]; } }int main(){int ncase=0;char str[100],str1[100],str2[100];double rate;while(scanf("%d",&n),n){co.clear();for(int i=1;i<=n;i++){scanf("%s",str);co[str]=i;}scanf("%d",&m);for(int i=0;i<m;i++){scanf("%s%lf%s",str1,&rate,str2);dis[co[str1]][co[str2]] = rate;}flag=0;dijkstra(1);printf("Case %d: ",++ncase);if(ans>1.0) printf("Yes\n");else printf("No\n");}return 0;}