UVA 436 Arbitrage (II)【floyd】

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=377

题意：硬币兑换，能不能兑换一圈使得手中的钱变多。
floyd传递闭包。

代码：

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <queue>
#include <math.h>
#include <map>
#include <string>

using namespace std;

int n, m;

map<string, int> MP;
double a[50][50];

int main()
{
    string s,y;
    double tmp;
    int cases = 1;
    while (cin >> n && n)
    {
        memset(a, 0, sizeof(a));
        MP.clear();
        for (int i = 1; i <= n; i++)
        {
            cin >> s;
            MP[s] = i;
        }
        for (int i = 1; i <= n; i++)
            a[i][i] = 1.0;
        cin >> m;
        for (int i = 1; i <= m; i++)
        {
            cin >> s >> tmp >> y;
            a[MP[s]][MP[y]] = tmp;
        }
        for (int k = 1; k <= n; k++)
            for (int i = 1; i <= n; i++)
                for (int j = 1; j <= n; j++)
                {
                    if (a[i][j] < a[i][k] * a[k][j])
                        a[i][j] = a[i][k] * a[k][j];
                }
        int ok = 0;
        for (int i = 1; i <= n; i++)
            if (a[i][i] > 1.0)
            {
                ok = 1;
                break;
            }
        printf("Case %d: ",cases++);
        if (ok) printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}