Arbitrage（最短路 + Bellman Ford + map）

Arbitrage

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14163		Accepted: 5957

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

 

    题意：

    给出 N（1 ~ 30），代表有 N 种钱币，后给出这 N 种钱币。后给出 M 条单向关系，代表前者兑换为后者的利率。问是否能使某种钱币升值。

 

    思路：

    最短路。同样也是Bellman ford，判断正环，利率可以不断相乘来累加，找到正环就能不断累加钱币，每种钱币都用字符串表示，转化为map对应匹配即可。

 

    AC：

#include <map>
#include <string.h>
#include <string>
#include <cstdio>
using namespace std;

typedef struct {
    int u,v;
    double w;
}node;

node edg[1000];
int ind,n;
double d[35];

int Bellman_ford() {
    memset(d,0,sizeof(d));
    d[1] = 1;
    for(int i = 1;i < n;i++)
        for(int j = 0;j < ind;j++) {
        int x = edg[j].u,y = edg[j].v;
        double r = edg[j].w;
        if(d[y] < d[x] * r) d[y] = d[x] * r;
    }

    for(int i = 0;i < ind;i++) {
        int x = edg[i].u,y = edg[i].v;
        double r = edg[i].w;
        if(d[y] < d[x] * r) return 1;
    }

    return 0;
}

int main() {
    //freopen("test.in","r",stdin);
    int time = 0;
    map<string,int> m;
    while(~scanf("%d",&n) && n) {
        int k;
        ind = 0;
        time++;
        for(int i = 1;i <= n;i++) {
            char str[100];
            scanf("%s",str);
            m[str] = i;
        }

        scanf("%d",&k);
        while(k--) {
            char st[100],en[100];
            double val;
            scanf("%s%lf%s",st,&val,en);
            edg[ind].u = m[st];
            edg[ind].v = m[en];
            edg[ind].w = val;
            ind++;
        }

        printf("Case %d: ",time);
        if(Bellman_ford())  puts("Yes");
        else    puts("No");
    }
    return 0;
}