2240:Arbitrage

总时间限制: 1000ms 内存限制: 65536kB
描述
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.
输入
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.
输出
For each test case, print one line telling whether arbitrage is possible or not in the format “Case case: Yes” respectively “Case case: No”.
样例输入
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
样例输出
Case 1: Yes
Case 2: No

最短路的变形，把每种货币看成节点，将汇率看成边的长度，计算时路径不是加法变成乘法啦；
使用的floyd算法，详见：http://blog.csdn.net/zhongkeli/article/details/8832946

#include <iostream>#include<stdio.h>#include<string>#include<string.h>#include<map>#define N 100#define inf 0#define MAX 10000using namespace std;double dis[N][N];int main(){    int n,i,m,k,j;    map<string,int> node;    char s[MAX],y[MAX];    double dis1;    int cnt=0;    while(~scanf("%d",&n)&&n)    {        memset(dis,inf,sizeof(dis));        memset(s,'\0',sizeof(s));        memset(y,'\0',sizeof(y));        for(i=0; i<n; i++)        {            cin>>s;            node[s]=i;            dis[i][i]=1;        }        scanf("%d",&m);        for(i=0; i<m; i++)        {            cin>>s>>dis1>>y;            dis[node[s]][node[y]]=dis1;        }        for(k=0; k<n; k++)            for(i=0; i<n; i++)                for(j=0; j<n; j++)                {                    if(dis[i][k]*dis[k][j]>dis[i][j])///若通过中间某个节点后钱币变多则选取这条路径                        dis[i][j]=dis[i][k]*dis[k][j];                }        bool flag=false;        for(i=0; i<n; i++)        {            if(dis[i][i]>1)//判断是否存在自己到自己大于1的路径            {                flag=true;                break;            }        }        printf("Case %d: %s\n",++cnt,flag==true?"Yes":"No");        node.clear();    }    return 0;}