最短路(Floyd_Warshall) POJ 2240 Arbitrage

题目传送门

/*
    最短路：Floyd模板题
        只要把+改为*就ok了，热闹后判断d[i][i]是否大于1
    文件输入的ONLINE_JUDGE少写了个_，WA了N遍:)
*/
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <string>
#include <map>
#include <cmath>
#include <vector>
#include <set>
#include <queue>
using namespace std;

const int MAXN = 1e6 + 10;
const int INF = 0x3f3f3f3f;
double d[33][33];

void Floyd_Warshall(int n)
{
    for (int k=1; k<=n; ++k)
    {
        for (int i=1; i<=n; ++i)
        {
            for (int j=1; j<=n; ++j)
            {
                if (d[i][j] < d[i][k] * d[k][j])
                {
                    d[i][j] = d[i][k] * d[k][j];
                }
            }
        }
    }

    for (int i=1; i<=n; ++i)
    {
        if (d[i][i] > 1)
        {
            puts ("Yes");    return ;
        }
    }

    puts ("No");    return ;
}

int main(void)        //POJ 2240 Arbitrage
{
    #ifndef ONLINE_JUDGE
    freopen ("F.in", "r", stdin);
    #endif

    int n, num;    int cas = 0;
    while (cin >> n && n)
    {
        for (int i=1; i<=n; ++i)
        {
            for (int j=1; j<=n; ++j)
            {
                if (i == j)        d[i][j] = 1;
                else    d[i][j] = 0;
            }
        }

        map<string, int> m;
        string s, s1, s2;
        for (int i=1; i<=n; ++i)
        {
            cin >> s;
            m[s] = i;
        }

        cin >> num;
        for (int i=1; i<=num; ++i)
        {
            double w;
            cin >> s1 >> w >> s2;
            d[m[s1]][m[s2]] = w;
        }

        cout << "Case " <<  ++cas << ": ";
        Floyd_Warshall (n);
    }

    return 0;
}