POJ

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

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

      题目大意：一个捣鼓钱的，靠汇率中间价挣钱，问是否能挣钱（置换回原来的钱）

  首先要将字符串换成数字啦，这是肯定的，（我一开始做啦）。。然后开始纠结是用最小生成树还是最短路，最后还是最短路啦，因为最小生成树在这题上没有意义。

不过呢，我还是想麻烦了一点。后来才知道直接floyd就好了。。。。

如果i==j时map[i[[j]=1，有时候其他的改一点也没关系，因为只是比较大小，，不过还是这样好理解。

不说了，上代码：

#include <stdio.h>#include <string.h>#define INF 0x3f3f3f3fdouble map[50][50];int n,m;void input(){    char s[50][100],a[100],b[100];    int i,k,j;    double c;    for(i=0; i<=n; i++)        for(j=0; j<=n; j++)            if(i==j)                map[i][j]=1; //最好是一            else                map[i][j]=-1;//这里不能是最大值啦，因为要求“最长路”，所以越小越好，。    for(i=1; i<=n; i++)        scanf("%s",s[i]);    scanf("%d\n",&m);    for(i=1; i<=m; i++)    {        scanf("%s %lf %s",a,&c,b);        for(j=1; j<=n; j++)                   //换数字            if(!strcmp(s[j],a)) break;        for(k=1; k<=n; k++)            if(!strcmp(s[k],b)) break;        map[j][k]=c;    }}void floyd(){    int i,j,k;    for(k=1; k<=n; k++)        for(i=1; i<=n; i++)            for(j=1; j<=n; j++)                if(map[i][j]<map[i][k]*map[k][j])                    map[i][j]=map[i][k]*map[k][j];}int main(){    int f=1,i;    while(~scanf("%d\n",&n) && n)    {        input();        floyd();        printf("Case %d: ",f++);        for(i=1; i<=n; i++)            if(map[i][i]>1) break;      //大于一就yes。        if(i>n) printf("No\n");        else printf("Yes\n");    }    return 0;}