poj2240floyd算法

Arbitrage

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11589		Accepted: 4866

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

找到一个圈，使得圈上边的权值之积>1，弗洛伊德算法可以求最短路径，改一下，也可以求最大乘积，如果最后出现某个w[i][i]>1，就说明存在这样的圈

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <string>
#define INF    1e8
using namespace std;
int n, m;
string s[30];
double w[30][30];
int getidx(const string &name)
{
    for(int i = 0; i < n; i++){
        if(s[i] == name){
            return i;
        }
    }
    printf("error\n");
    return -1;
}
void floyd()
{
    for(int k = 0; k < n; k++){
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                if(w[i][j] < w[i][k] * w[k][j]){
                    w[i][j] = w[i][k] * w[k][j];
                }
            }
        }
    }
}
int main(int argc, const char *argv[])
{
    int cnt = 1;
    while(cin >> n && n != 0){
        for(int i = 0; i < n; i++){
            cin >> s[i];
        }
        memset(w, 0, sizeof(w));//init
        for(int i = 0; i < n; i++){
            w[i][i] = 1;
        }
        cin >> m;
        string tmp1, tmp2;
        double tmpw;
        for(int i = 0; i < m; i++){
            cin >> tmp1 >> tmpw >> tmp2;
            w[getidx(tmp1)][getidx(tmp2)] = tmpw;
        }
        floyd();
        int i;
        for(i = 0; i < n; i++){
            if(w[i][i] > 1){
                printf("Case %d: Yes\n", cnt);
                break;
            }
        }
        if(i == n){
            printf("Case %d: No\n", cnt);
        }
        cnt++;
    }
    return 0;
}

其实我还没有完全理解，可是就AC了……(用G++超时，然后跑到discuss版看了一下，有人说C++更快，果然，79MS通过……怎么差这么多)

floyd算法用的是算法导论25章习题25.2-4的改进版（空间从n^3减少到n^2）